Every Linear Order Isomorphic to Its Cube Is Isomorphic to Its Square

EVERY LINEAR ORDER ISOMORPHIC TO ITS CUBE IS ISOMORPHIC TO ITS SQUARE

GARRETT ERVIN

Abstract. In 1958, Sierpi´nskiasked whether there exists a linear order X that is isomorphic to its lexicographically ordered cube but is not isomorphic to its square. The main result of this paper is that the answer is negative. More generally, if X is isomorphic to any one of its ﬁnite powers Xn, n > 1, it is isomorphic to all of them.

1. Introduction Suppose that (C, ×) is a class of structures equipped with a product. In many cases it is possible to find an infinite structure X ∈ C that is isomorphic to its own square. If X2 =∼ X, then by multiplying again we have X3 =∼ X. Determining whether the converse holds for a given class C, that is, whether X3 =∼ X implies X2 =∼ X for all X ∈ C, is called the cube problem for C. If the cube problem for C has a positive answer, then C is said to have the cube property. The cube problem is related to three other basic problems concerning the multiplication of structures in a given class (C, ×). 1. Does A × Y =∼ X and B × X =∼ Y imply X =∼ Y for all A, B, X, Y ∈ C? Equivalently, does A × B × X =∼ X imply B × X =∼ X for all A, B, X ∈ C? 2. Does X2 =∼ Y 2 imply X =∼ Y for all X,Y ∈ C? 3. Does A × Y =∼ X and A × X =∼ Y imply X =∼ Y for all A, X, Y ∈ C? Equivalently, does A2 × X =∼ X imply A × X =∼ X for all A, X ∈ C? The first question is called the Schroeder-Bernstein problem for C, and if it has a positive answer, then C is said to have the Schroeder-Bernstein property. The second question is called the unique square root problem for C, and if its answer is positive, then C has the unique square root property. Taken together, the first two questions are sometimes called the Kaplansky test problems, after Irving Kaplansky who posed them in [7] as a heuristic test for whether a given class of abelian groups (under the direct product) has a satisfactory structure theory (“I believe their defeat is convincing evidence that no reasonable invariants exist”). Tarski [14] had posed them previously for the class of Boolean algebras. All three questions are listed in Hanf’s seminal paper [4] on products of Boolean algebras. We will refer to Question 3 as the weak Schroeder-Bernstein problem for C, and the corresponding property as the weak Schroeder-Bernstein property. A negative solution to Question 3 obviously gives a negative solution to Question 1. If the product for C is commutative, it gives a negative solution to Question 2 as well. If the cube problem for C has a negative solution, that is, if there is an X ∈ C that is isomorphic to its cube but not to its square, then all three questions have a negative solution, even without assuming commutativity of the product. In practice, it is often by constructing such an X that these three problems are solved. 1 2 GARRETT ERVIN

If the class C does not contain any infinite structure isomorphic to its cube, then the cube property holds trivially. When the cube property does not hold trivially, it usually fails. The first result in this direction is due to Hanf, who constructed in [4] a Boolean algebra that is isomorphic to its cube but not its square. Tarski [15] and Jónsson[6] immediately adapted Hanf’s result to show the failure of the cube property for the class of semigroups, the class of groups, the class of rings, and various other classes of algebraic structures. Hanf’s example, and consequently many of those produced by Tarski and Jónsson,is of size continuum, and for some time it was open whether there were countable examples witnessing the failure of the cube property for these various classes. In 1965, Corner showed in [1] that indeed there exists a countable (torsion-free, abelian) group G isomorphic to G3 but not G2. Later, Jones [5] showed that it is even possible construct a finitely generated (necessarily non-abelian) group isomorphic to its cube but not its square. In 1979, Ketonen [8] solved the so-called Tarski cube problem by producing a countable Boolean algebra isomorphic to its cube but not its square. Throughout the 1970s and 1980s, Trnkovásolved the cube problem negatively for many different classes of topological spaces and relational structures, including the class of graphs under several different notions of graph product [19]. Her topological results are summarized in [18]. Answering a question of Trnková,Orsatti and Rodino showed in [11] that there is even a connected topological space homeomorphic to its cube but not its square. Koubek, Neˇsetˇril,and Rödl[9] showed that the cube property fails for the class of partial orders, as well as for other classes of relational structures. More recently, Eklof and Shelah [2] constructed an ℵ1-separable group isomorphic to its cube but not its square, and Gowers [3] constructed a Banach space linearly homeomorphic to its cube but not its square. On the other hand, there are rare instances when the cube property holds non- trivially. It holds for the class of sets under the cartesian product, since any set in bijective correspondence with its cube is either infinite, empty, or a singleton, and hence in bijection with its square. This is immediate if one assumes the axiom of choice, but it can be proved without the axiom of choice using the Schroeder- Bernstein theorem. Similarly easily, the cube property holds for the class of vector spaces (over a fixed field) under the direct product. Less trivially, the cube property holds for the class of countably complete Boolean algebras. This follows from the Schroeder-Bernstein theorem for such algebras. Trnková[16] showed that the cube property also holds for the class of countable metric spaces (where isomorphism means homeomorphism), as well as for closed subspaces of Cantor space [17]. Koubek, Neˇsetˇril,and Rödlshowed in [9] that the cube property holds for the class of equivalence relations. It is worth noting that for all of these classes, it is actually possible to establish the stronger Schroeder-Bernstein property. In his 1958 book Cardinal and Ordinal Numbers [13], Sierpińskiposed the cube problem (although he does not use the term) for the class (LO, ×lex) of linear orders under the lexicographical product. On page 232, he writes, “We do not know so far ... any type α such that α = α3 6= α2.” Here, “type” means linear order type, and the ordering on the cartesian powers α2 and α3 is the lexicographical ordering1. Although the cube problem has been solved

1Sierpińskiactually ordered these powers anti-lexicographically, though in this paper we will use the lexicographical ordering. This does not change the problem. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 3 for many other classes of structures, Sierpiński’squestion has remained open. One major difference in this version of the cube problem is that, unlike the products for the other classes so far discussed, the lexicographical product of linear orders is not commutative. Though he does not make a conjecture in his book, his language suggests that Sierpińskiexpected that such an α exists, that is, that the cube property fails for (LO, ×lex). He was already aware of examples of linear orders witnessing the failure of the unique square root property and (the right-sided and left-sided versions of) the Schroeder-Bernstein property. The main result of this paper is that in fact the cube property holds for (LO, ×lex). This appears as Theorem 4.14 below. Main Theorem. If X is a linear order and X3 =∼ X, then X2 =∼ X. More generally, if Xn =∼ X for some n > 1, then X2 =∼ X. Thus the cube property holds for the class of linear orders despite the fact that the Schroeder-Bernstein property and unique square root property fail. We will show in Section 6 that even the weak Schroeder-Bernstein property fails for (LO, ×lex). In this sense, the cube property is closer to failing for (LO, ×lex) than it is for the other classes for which it is known to hold. In the remainder of this paper we prove the main theorem as well as some related results. In Section 2, we define the necessary notation and terminology, and give some examples of countable linear orders isomorphic to their own squares. We also give the easy proof that the cube property holds for linear orders with both a top and bottom point. In Section 3, we prove a representation theorem for linear orders X that are invariant under left multiplication by a fixed order A, that is, that satisfy the isomorphism A×X =∼ X. It turns out that such orders can essentially be represented as unions of tail-equivalence classes in the lexicographically ordered tree Aω. This theorem can be generalized to characterize the isomorphism An × X =∼ X for any n ≥ 1, and in particular A2 × X =∼ X. It is then possible to write down a sufficient condition, namely the existence of a parity-reversing automorphism of Aω, for the implication A2 × X =∼ X =⇒ A × X =∼ X to hold for every X. The combinatorial heart of the proof is contained in Section 4, where parity- reversing automorphisms of Aω are constructed for various orders A, including all countable orders. These maps are built using generalized versions of the classical Schroeder-Bernstein bijection. In this context, the isomorphism X3 =∼ X can be rewritten X2 × X =∼ X. The results of Sections 3 and 4 then give that if Xω has a parity-reversing automorphism, then we must also have X × X =∼ X (i.e. X2 =∼ X). We will show at the end of Section 4 that if X3 =∼ X, then indeed Xω has a parity-reversing automorphism, completing the proof of the cube property. The proof is then generalized to give Xn =∼ X =⇒ X2 =∼ X for any linear order X and n > 1. In Section 5, we show that the main theorem is not vacuous by illustrating a general way of constructing orders X such that Xn =∼ X for a fixed n > 1. Such orders can be arranged to be of any cardinality. In Section 6 we will show that there exists an order A such that Aω does not have a parity-reversing automorphism, and as a consequence we will be able to construct a counterexample to the weak Schroeder-Bernstein property for (LO, ×lex). In Section 7, several related problems concerning the multiplication of linear orders are discussed. The paper is for the most part self-contained. General background on linear orders can be found in [12]. 4 GARRETT ERVIN

I am indebted to Justin Moore for carefully reading an earlier version of this paper and oﬀering a number of useful suggestions.

2. Preliminaries

2.1. Terminology. A linear order is a pair (X, isomorphisms factor through products and replacements. If X and Y are isomorphic orders, as witnessed by an isomorphism f : X → Y , then for any order Z, the products XZ and YZ are isomorphic, as witnessed by the map (x, z) 7→ (f(x), z). Similarly, if we have collections of orders Ix, x ∈ X, and Jy, y ∈ Y so that Ix = Jf(x) for all x ∈ X, we will also have that the replacements X(Ix) and Y (Jy) are isomorphic, as witnessed by the map (x, i) 7→ (f(x), i).

2.2. Examples of countable orders isomorphic to their squares. To begin, we give some examples of countable orders that are isomorphic to their own squares. In Section 5 we will construct uncountable examples.

2 P The order X(Ix) is sometimes called an ordered sum and denoted x∈X Ix. However, in this paper we deal with cases in which X(Ix) behaves more like a product-by-X than a sum-over-X, so this notation is more convenient. 6 GARRETT ERVIN

Our examples here rely on Cantor’s theorem characterizing the order types of countable dense linear orders, as well as a generalization due to Skolem. Theorem. (Cantor) If X and Y are countable dense linear orders without endpoints, then X is isomorphic to Y . Thus every countable dense linear order without endpoints is isomorphic to the set of rationals Q in their usual order. Theorem. (Skolem) Fix some k, 1 ≤ k ≤ ω. Let X,Y be countable dense linear S orders without endpoints. Fix a partition X = Xi such that each Xi is dense S i 1, then X has the same form as the order above: ∼ X = Q(Iq), where for each q there is a dense set of p ∈ Q such that Iq = Ip. It follows, as above, that X2 =∼ X and hence Xm =∼ X for any m > 1. Of course, our main theorem is that Xn =∼ X =⇒ X2 =∼ X holds in general, but in the countable, no-endpoints case we have a complete classification: Xn =∼ X for some n > 1 if and ∼ only if X = Q(Iq). As we observed, such an order is not only invariant under left multiplication by itself, but by any countable order L. 2.3. A solution to the cube problem when X has both endpoints. It turns out that it is easy to prove Xn =∼ X =⇒ X2 =∼ X for linear orders X with both a left and right endpoint. The proof uses the following theorem of Lindenbaum, THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 7 that could be called the Schroeder-Bernstein theorem for linear orders. Although the proof for the general case is substantially harder, a generalized version of Lin- denbaum’s theorem does play an important role. Theorem. (Lindenbaum) Suppose X,Y are linear orders. If X is isomorphic to an initial segment of Y , and Y is isomorphic to a final segment of X, then X =∼ Y . Proof. Suppose f : X → Y is an isomorphism of X onto an initial segment of Y and g : Y → X is an isomorphism of Y onto a final segment of X. Then f, g are in particular injections. Let h : X → Y be the bijection constructed out of f, g as in the classical proof of the Schroeder-Bernstein theorem. Then the hypotheses guarantee that this bijection is order-preserving. Corollary 2.1. If X is a linear order with both a left endpoint and right endpoint, and Xn =∼ X for some n > 1, then X2 =∼ X. Proof. Denote the minimal element of X by 0 and the maximal element by 1. Then X2 contains an initial segment isomorphic to X, namely the interval of points of the form (0, ·). Since Xn =∼ X, it follows that X2 contains an initial segment isomorphic to Xn. But the interval of consisting of points (1, 1,..., 1, ·, ·) with n − 2 leading 1s is a final segment of Xn isomorphic to X2. By Lindenbaum’s theorem, X2 =∼ Xn. 2 ∼ Hence X = X. Thus any linear order with both endpoints that is isomorphic to its cube is isomorphic to its square, solving the cube problem in this case. The simplicity of the proof suggests that there may be a simple proof when X has either no endpoints or one endpoint. But the proof uses crucially that X has both endpoints, and there does not seem to be an adaptation to the other cases. The isomorphisms constructed in Section 4 to take care of the other cases do make use of Schroeder-Bernstein style maps, but using them requires an understanding of the structure of orders X that satisfy isomorphisms of the form AnX =∼ X.

3. Characterizing orders X such that AnX =∼ X 3.1. The isomorphism AX =∼ X. Throughout the rest of the paper, when it is not otherwise specified, A simply refers to some fixed order. Our first task is to understand which orders X are isomorphic to AX. We begin with some examples. ω Example 3.1. Let X = A . The map from AX to X defined by (a, (a0, a1,...)) 7→ ∼ (a, a0, a1,...) is order-preserving and bijective. Thus AX = X. Since it will appear often, let us denote the “flattening map” in the example by ω ω fl : A × A → A , where fl(a, (a0, a1,...)) = (a, a0, a1,...). Example 3.2. Choose a distinguished element in A and label it 0. Let X be the suborder of Aω consisting of sequences that are eventually 0. Then AX =∼ X. Again the isomorphism is given by fl, restricted here to AX. Similarly, if X is the subset of Aω consisting of eventually constant sequences, then AX is isomorphic to X under the same map. These examples can be generalized. The relevant notion is that of tail-equivalence of sequences in Aω. In what follows, A<ω denotes the set of finite sequences of elements of A, including the empty sequence. We do not put any order structure on this set. If r is a finite sequence and u is a finite or infinite sequence, then ru denotes the sequence obtained by concatenating r with u. 8 GARRETT ERVIN

Definition 3.3. Two sequences u, v ∈ Aω are called tail-equivalent if there exist sequences r, s ∈ A<ω (not necessarily of the same length) and a sequence u0 ∈ Aω such that u = ru0 and v = su0. If u and v are tail-equivalent, we write u ∼ v. In dealing with sequences of elements of A, the letters a, b, . . . will usually denote single elements of A, whereas r, s, . . . will denote finite sequences, and u, v, . . . will denote infinite sequences. If u = ru0 for some finite sequence r, then u0 is called a tail-sequence of u and r is called an initial sequence of u. In an abuse, no distinction is made between elements of A and sequences of length 1, so that au refers to the sequence with first entry a ∈ A followed by the tail-sequence u ∈ Aω. The length ω of a finite sequence r is denoted |r|. Given u ∈ A , ui refers to the ith entry of u, and u n = (u0, u1, . . . , un−1) denotes the initial sequence of the first n entries of u. If u ∼ v, so that u = ru0 and v = su0 for some r, s, u0, the pair ru0 and su0 is called a meeting representation of u and v. Meeting representations are not unique: if u = ru0, v = su0, and a is the first entry of u0, so that u0 = au00, then letting r0 = ra, s0 = sa we have that u = r0u00 and v = s0u00, giving a different representation. Usually, unzipping along a tail- sequence like this is the only way of generating distinct representations. However, if u0 is an eventually periodic sequence there are other ways. We will return to this issue later. It is easy to see that tail-equivalence is an equivalence relation on Aω. The equivalence class of u is denoted [u]. It consists exactly of those elements in Aω of the form ru0, where r ∈ A<ω and u0 is a tail-sequence of u. The tail-equivalence classes are the smallest subsets of Aω that are “invariant under left multiplication by A” in the following sense. If [u] is a tail-equivalence class, then viewing it as a suborder of Aω, we may form the product A[u] = {(a, v): a ∈ A, v ∈ [u]}. Then fl[A[u]] = [u], that is, the flattening map (a, v) 7→ av witnesses the isomorphism A[u] =∼ [u]. To check this, note that if v ∈ [u], then av ∼ v and hence av ∈ [u] for any a ∈ A. Hence fl[A[u]] ⊆ [u]. And if v ∈ [u], say 0 v = (v0, v1,...), then v = (v1, v2,...) is tail-equivalent to v and hence also in [u]. 0 But then v0v = v is in the image of A[u], giving the reverse containment. On the other hand, if X ⊆ Aω and fl[AX] = X, then X is a union of tail- −1 equivalence classes. For, if u = (u0, u1,...) is in X, then fl (u) = (u0, (u1,...)) is in AX and therefore the tail-sequence (u1, u2,...) is in X. By induction, any tail- sequence u0 of u is in X. Then for any a ∈ A, we must have (a, u0) ∈ AX and hence fl((a, u0)) = au0 ∈ X. By induction, for any finite sequence r we have ru0 ∈ X. Thus [u] ⊆ X, giving the claim. We have proved the following proposition. Proposition 3.4. Fix a suborder X ⊆ Aω. Then fl[AX] = X if and only if X is a union of tail-equivalence classes. This fact can be formulated dynamically. We may view A<ω as the free semigroup on the alphabet A, where the semigroup operation is concatenation. This semigroup naturally acts on Aω by concatenation on the left. The tail-equivalence classes are the smallest subsets of Aω that are closed under this action by A<ω. Our aim now is to find the most general form of an order X such that AX =∼ X. Proposition 3.4 gives immediately that if X is any suborder of Aω that is a union of tail-equivalence classes, then AX is isomorphic to X under the flattening map (a, u) 7→ au. Notice that the X’s appearing in Examples 3.1 and 3.2 are of this form: Aω is the union of all of its tail-equivalence classes, and the set of eventually THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 9

0 sequences is exactly the single tail-equivalence class of the constant sequence (0, 0,...). The proposition actually yields substantially more general examples. Suppose ω that for every u ∈ A we fix an order Iu, with the restriction that if u ∼ v then ω ∼ Iu = Iv. Let X = A (Iu). Then AX = X, and the isomorphism is witnessed by the flattening isomorphism on the left two coordinates. To see this, rewrite ω ω AX = A × A (Iu) as (A × A )(J(a,u)), where J(a,u) = Iu for all a ∈ A. Then J(a,u) = Iau as well, since au ∼ u. This is the same as saying that the interval ω consisting of points (a, u, ·) in (A × A )(J(a,u)) is isomorphic to the interval of ω points (au, ·) in A (Iu). Thus the map (a, u, x) 7→ (au, x) makes sense, and defines ω ω an order isomorphism of (A × A )(J(a,u)) with A (Iu), that is, of AX with X. Letting I[u] denote the single order Iv for all v ∈ [u], we may denote X by ω ω A (I[u]). Our previous examples were actually of this form: if X ⊆ A is a union ω of tail-equivalence classes then X may be written as A (I[u]), where I[u] = 1 if u ∈ X, and otherwise I[u] = ∅. Of course, since the I[u] may be chosen arbitrarily, there are many other examples besides these. Here is a concrete one. Let Z denote the integers in their usual order, and form the product Zω (this order is isomorphic to the irrationals). Each tail-equivalence class [u] ⊆ Zω is a countable dense subset of Zω, hence the number of classes is ω ω 2 . Enumerate them as {Cα : α < 2 }. Let X be the order obtained by replacing every point in the αth class with the ordinal α, i.e. if [u] = Cα let I[u] = α, and let ω X = Z (I[u]). This order may be visualized as densely many copies of each ordinal less than 2ω interspersed with one another. By our observations above, ZX =∼ X. We shall now prove that this form is general: any order X such that AX =∼ X ω must be of the form A (I[u]).

Theorem 3.5. Fix an order X. Then AX =∼ X if and only if X is isomorphic to ω an order of the form A (I[u]).

ω Proof. We have already seen that if X is of the form A (I[u]) then X is isomorphic to AX. So assume that AX =∼ X, as witnessed by an isomorphism f : AX → X. For every a ∈ A, let aX denote the set of pairs in AX with first coordinate a. Let Xa denote f[aX]. Each aX is an interval in AX isomorphic to X, and hence each Xa is an interval in X isomorphic to X. These intervals are disjoint and cover X, and if a < b, then the interval Xa lies to the left of Xb, that is, Xa < Xb. We think ∼ of AX = X as saying that X can be split into A-many copies of X, and Xa as being the ath copy of X within X. Let fa : X → Xa be the isomorphism defined by fa(x) = f(a, x). Given one of the intervals Xb and some a ∈ A, we may think of the image fa[Xb] as the bth copy of X within Xa. Denote this interval by X(a,b). Then fa ◦ fb is an isomorphism of X with X(a,b). Extending this, given a finite sequence <ω r = (a0, a1, . . . , an) ∈ A , define Xr as fa0 ◦ fa1 ◦ ... ◦ fan [X]. Denote the isomorphism fa0 ◦ fa1 ◦ ... ◦ fan : X → Xr by fr. <ω It is immediate that for any r, s ∈ A , we have fr[Xs] = Xrs, and conversely, −1 fr [Xrs] = Xs. If t extends r, then Xt is a subinterval of Xr. If neither one of r, t extends the other, and they differ first in the nth place with rn < tn, then Xr < Xt in X. 10 GARRETT ERVIN

ω Now, given an inﬁnite sequence u ∈ A , deﬁne Iu to be the interval obtained by taking the natural nested intersection: \ Iu = Xun. n∈ω

For a given u, the interval Iu need not be isomorphic to X. Indeed, Iu may be ω empty. However, every x ∈ X is in some Iu. Further if u < v in A , then for some n we have u n < v n. Hence Xun < Xvn in X, and so Iu < Iv in X. Thus we have that the Iu in fact partition X, and further that they respect the ∼ ω lexicographical ordering of their indices. This shows X = A (Iu). ∼ 0 0 It remains to show that for u ∼ v, we have Iu = Iv. If u = ru and v = su , then if we begin in Xr and Xs (each copies of X) and descend in each interval along 0 the same nested sequence of subintervals prescribed by u , we arrive at Iu and Iv, ∼ −1 respectively. Hence, Iu = Iv. The isomorphism is given by fs ◦ fr , restricted to Iu. To see this explicitly, note that we have \ I = X 0 , u r(u n) n −1 where r(u n) means r concatenated with (u n). Thus, since fs ◦ fr : Xr → Xs is an isomorphism, we have

−1 −1 T f ◦ f [I ] = f ◦ f [ X 0 ] s r u s r n r(u n) T −1 = fs ◦ fr [Xr(u0 n)] Tn = X 0 n s(u n) = Isu0 = Iv −1 ∼ That is, fs ◦ fr witnesses Iu = Iv, as claimed. ∼ Since u ∼ v implies Iu = Iv, we may write I[u] for the single order type of Iv for ∼ ω all v ∈ [u]. We have showed X = A (I[u]). This theorem plays a basic role in the remainder of the paper, and some discus- ω sion is warranted. We have seen that if X = A (I[u]) for some collection of orders I[u], then there is a natural isomorphism f : AX → X defined by (a, u, x) 7→ (au, x). That is, f is the flattening map on the first two coordinates and the identity on the last. We write f = (fl, id). ω Even if we do not have identity, but only an isomorphism F : X → A (I[u]), there is still a natural isomorphism f : AX → X, namely f = F −1 ◦(fl, id)◦(id,F ). This is the same flattening map, up to the relabeling F . On the other hand, suppose we are given an isomorphism f : AX → X. The proof of Theorem 3.5 shows that we can use this f to build an isomorphism F from ω X onto an order of the form A (I[u]). In fact the proof shows more: if we view f, ω ω via the relabeling F , as an isomorphism of A × A (I[u]) with A (I[u]), then f is just the flattening map on the first two coordinates. (It need not be true that f is the identity on the last coordinate, but we shall ignore this subtlety.) In this sense, there is a correspondence between isomorphisms f : AX → X ω ω and labelings F : X → A (I[u]). Given a labeling of X as A (I[u]), we have the natural flattening isomorphism between AX and X, and given an isomorphism ω f : AX → X, we can relabel X as A (I[u]) in such a way that f is just the flattening isomorphism. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 11

3.2. The isomorphism A2X =∼ X. Our next goal is to characterize, for a ﬁxed n > 1, those X for which AnX =∼ X. We focus here on the case when n = 2, but the work easily generalizes. A list of the corresponding deﬁnitions and results for the n case is provided later in the section. In a sense, we have already succeeded in getting a characterization. By Theorem 2 ∼ 2 ω 3.5, A X = X if and only if X is isomorphic to an order of the form (A ) (I[u]), where now [u] denotes a tail-equivalence class in (A2)ω. But (A2)ω is naturally ω isomorphic to A under the map ((a0, a1), (a2, a3),...) 7→ (a0, a1, a2, a3,...). It is useful to characterize such an X as a replacement of Aω instead of as a replacement of (A2)ω.

Definition 3.6. Two sequences u, v ∈ Aω are called 2-tail-equivalent if there exist r, s ∈ A<ω with |r| ≡ |s| (mod 2) and a sequence u0 ∈ Aω such that u = ru0 and 0 v = su . If u and v are 2-tail-equivalent, we write u ∼2 v. The definition of 2-tail-equivalence is identical to that of tail-equivalence, except 0 0 that in the meeting representation u = ru , v = su witnessing u ∼2 v, the lengths of r and s are required to have the same parity. Reflexivity and symmetry of the relation ∼2 are immediate, and transitivity can be checked. Hence 2-tail-equivalence is an equivalence relation. The ∼2-equivalence class of u is denoted [u]2. We will need to see precisely how tail-equivalence classes are related to 2-tail- equivalence classes. If u ∼2 v, then certainly u ∼ v as well. Hence the 2-tail- equivalence relation refines the tail-equivalence relation, and every ∼-class [u] splits into some number of ∼2-classes. Clearly in fact, it splits into at most two such classes: if v ∼ u as witnessed by u = ru0 and v = su0, then either |s| ≡ |r| (mod 2) or |s| ≡ |r| + 1 (mod 2). In the first case we have v ∼2 u, and in the second, for any a ∈ A, we have |s| ≡ |ar| (mod 2), and hence v ∼2 au. Thus [u] = [u]2 ∪ [au]2 for any fixed a ∈ A. The classes [u]2 and [au]2 are either equal or disjoint, depending on whether au ∼2 u or not. If au ∼2 u, we have [u] = [u]2; otherwise [u] splits into the two disjoint classes [u]2 and [au]2. 0 0 Usually, au 6∼2 u. Certainly the obvious representation u = ru , au = su , 0 where r = ∅, s = a, and u = u does not witness au ∼2 u. Most of the time this obvious representation is the only one, up to unzipping along the tail-sequence. However, if u is an eventually periodic sequence then it is possible to get truly distinct representations, and in some of these cases we have au ∼2 u. For any finite sequence s ∈ A<ω, let s denote the sequence sss . . . ∈ Aω. Define a sequence u ∈ Aω to be eventually periodic if there exist r, s ∈ A<ω such that u = rsss . . . = rs. Such a sequence is said to be repeating in s. If r = ∅ we say u is simply periodic. If n is the shortest possible length of a sequence in which u repeats, we say u has period n. To prove the following proposition, we use the fact that if n is the period of u and we have a “periodic representation” of u as u = rs, then n divides |s|.

ω Proposition 3.7. Fix elements u ∈ A and a ∈ A. Then au ∼2 u if and only if u is eventually periodic and the period of u is odd.

Proof. Suppose u is repeating in s, and |s| is of minimal length, i.e. |s| = period of u. Assume ﬁrst that |s| ≡ 1 (mod 2). We have u = rs for some r ∈ A<ω. Let u0 = s be the periodic tail-sequence, and let t = ars. Then since |s| is odd, we have 12 GARRETT ERVIN

|t| = |ars| ≡ |r| (mod 2). Thus the meeting representation u = ru0 and au = tu0 witnesses au ∼2 u. 0 0 0 Conversely, suppose au ∼2 u, as witnessed by u = ru and au = (ar )u , with |r| ≡ |ar0| (mod 2). Hence |r| 6≡ |r0| (mod 2). Since both r and r0 are initial sequences of u, it must be that one is an initial sequence of the other. Assume r0 is an initial sequence of r, that is, r = r0s for some s ∈ Aω. Then since r, r0 are inequivalent modulo 2, it must be that |s| is odd. Now, on one hand u = ru0 = r0su0, and on the other u = r0u0. Hence u0 = su0. The only way this is possible is if u0 = s. Therefore u = rs is an eventually periodic sequence. It may be that the period of u is shorter than |s|, but since |s| is odd, the period of u must be odd as well. The 0 case when r is an initial sequence of r is similar.

Thus [u] = [u]2 if and only if u is eventually periodic and of odd period. Since they play an especially important role in what follows, we emphasize that for sequences u of period 1 we have [u] = [u]2 = [au]2. For sequences u that are not eventually periodic we have [u]2 ∩ [au]2 = ∅, where a is any fixed element of A. Just as the ∼-classes [u] were the smallest suborders of Aω invariant under left ω multiplication by A, the ∼2-classes [u]2 are the smallest suborders of A invariant 2 2 ω ω under left multiplication by A . Let fl2 : A × A → A denote the “2-flattening isomorphism” defined by (a, b, u) 7→ abu. ω 2 Proposition 3.8. Fix a suborder X ⊆ A . Then fl2[A X] = X if and only if X is a union of ∼2-classes. Proof. The proof is essentially identical to the proof of Proposition 3.4. There, the relevant fact was that au ∼ u for all a, u. Here, it is that abu ∼2 u for all a, b, u. ω 2 Just as fl sends A[u] isomorphically onto [u] for every u ∈ A , fl2 sends A [u]2 isomorphically onto [u]2. Less formally, we can express this by saying A[u] = [u] 2 ω and A [u]2 = [u]2 for all u ∈ A . A natural question is what happens to a given 2-tail-equivalence class [u]2 when it is multiplied by a single factor of A. ω Proposition 3.9. For all u ∈ A and a ∈ A we have fl[A[u]2] = [au]2.

Proof. If x ∈ fl[A[u]2], then x = bv for some b ∈ A and v ∼2 u. But then bv ∼2 au, 0 0 i.e. x ∼2 au. Conversely, if x ∼2 au then writing x = x0x we have that x ∼2 u and hence x ∈ fl[A[u]2].

Informally this says A[u]2 = [au]2. It follows that A[au]2 = [u]2, from which we 2 recover A [u]2 = [u]2. We can use Proposition 3.8 to produce orders X such that A2X =∼ X. For ω ω every class [u]2, ﬁx an order I[u]2 . Let A (I[u]2 ) denote the order A (Iu), where 2 ∼ if v ∼2 u, then Iv = Iu = I[u]2 . Then A X = X. The isomorphism is given by ((a, b), u, x) 7→ (abu, x). This map makes sense because both the interval consisting of points ((a, b), u, ·) in A2X and the interval of points (abu, ·) in X are of type

I[u]2 = I[abu]2 . ω 2 The form A (I[u]2 ) is general for orders X isomorphic to A X. Theorem 3.10. Fix an order X. Then A2X =∼ X if and only if X is isomorphic ω to an order of the form A (I[u]2 ). Proof. We have already showed the forward direction. Assume A2X =∼ X. Then 2 ω by Theorem 3.5 we have that X is of the form (A ) (J[w]), where [w] denotes a THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 13 tail-equivalence class in (A2)ω. Let F l :(A2)ω → Aω be the isomorphism defined by ((a0, a1), (a2, a3),...) 7→ (a0, a1, a2,...). It is straightforward to check that for w, x ∈ (A2)ω, we have w ∼ x (in the sense 2 ω ω of (A ) ) if and only if F l(w) ∼2 F l(x) (in the sense of A ). ω For every u ∈ A , define Iu to be JF l−1(u). Then Jw = IF l(w) for every w ∈ (A2)ω. Hence the isomorphism F l :(A2)ω → Aω lifts to give an isomorphism of 2 ω ω 2 ω (A ) (Jw) with A (Iu). Since w ∼ x in (A ) implies Jw = Jx, we have that ω ω u ∼2 v in A implies Iu = Iv. Thus X is of the form A (I[u]2 ). 3.3. When A2X =∼ X implies AX =∼ X. We now turn to the question of when the isomorphism A2X =∼ X implies AX =∼ X. Notice that if A2X =∼ X and ω we use Theorem 3.10 to decompose X as A (I[u]2 ), then there is no “obvious” isomorphism between AX and X. The natural guess for such an isomorphism would be the flattening map (a, u, x) 7→ (au, x). But the definition of this map may be meaningless: the interval consisting of tuples (a, u, ·) in AX is of type Iu = I[u]2 whereas the interval (au, ·) in X is of type I[au]2 . If u 6∼2 au, these orders may be distinct.

If it so happens that I[u]2 = I[au]2 for every u and a, then the ﬂattening map makes sense and witnesses AX =∼ X. Indeed if this is so we may denote the common ∼ ω order type of I[u]2 and I[au]2 by I[u], so that X = A (I[u]). But this need not be the case. For example, consider Zω and let v denote the sequence (1, 2, 3,...). Then v is not eventually periodic and hence not 2-tail- equivalent to 0v = (0, 1, 2,...). Let I[v]2 = 1 and I[0v]2 = 2. For all other ∼2-classes ω [u]2, let I[u]2 = ∅. Let X = Z (I[u]2 ). Then the map ((a, b), u, x) 7→ (abu, x) witnesses Z2X =∼ X, while the “map” (a, u, x) 7→ (au, x), which purports to send each copy of 1 in AX isomorphically onto a copy of 2 in X, and vice versa, is meaningless. In this sense it is not immediate that ZX =∼ X (though, in this case, this turns out to be true). The following proposition describes the precise relationship between AX and X when A2X =∼ X. It echoes Proposition 3.9. Proposition 3.11. Suppose X is an order such that A2X =∼ X, so that X is of ω ∼ ω ω the form A (I[u]2 ), and let Y = AX. Then Y = A (J[u]2 ), where for all u ∈ A and a ∈ A, we have J[u]2 = I[au]2 .

Proof. Deﬁne J[u]2 = I[au]2 as in the statement of the theorem, and assume for ω simplicity that X is not only isomorphic to but in fact equals A (I[u]2 ). Then the ω isomorphism between AX and A (J[u]2 ) is given by the ﬂattening map (a, u, x) 7→ (au, x). This is an isomorphism since each interval in AX consisting of points ω (a, u, ·) is of type I[u]2 and each interval (au, ·) in A (J[u]2 ) is of type J[au]2 =

I[aau]2 = I[u]2 as well. 2 ∼ Thus if A X = X, we obtain AX by interchanging the role of each I[u]2 with ∼ ω I[au]2 in the decomposition X = A (I[u]2 ). It is worth noting that in the case when

[u]2 = [au]2 = [u], no interchange is needed: in this case the orders I[u]2 , I[au]2 , and

J[u]2 , J[au]2 are all identical, whereas in general only the equalities J[u]2 = I[au]2 and I[u]2 = J[au]2 hold. The upshot of Proposition 3.11 is that, in the case where A2X =∼ X, if we can ω ﬁnd an order automorphism of A that maps each ∼2-class [u]2 onto [au]2, we can lift it to obtain an isomorphism of AX with X. 14 GARRETT ERVIN

Deﬁnition 3.12. An order automorphism f : Aω → Aω is called a parity-reversing ω automorphism (abbreviated p.r.a.) if f(u) ∈ [au]2 for every u ∈ A and a ∈ A. Equivalently, an order automorphism f of Aω is a p.r.a. if for every u ∈ Aω and a ∈ A the image of [u]2 under f is [au]2.

It follows that the image of [au]2 under a parity-reversing automorphism f is [u]2. Parity-reversing automorphisms are “idempotent on ∼2-classes” in this sense. Proposition 3.13. Suppose that Aω admits a parity-reversing automorphism. Then for every order X, if A2X =∼ X then AX =∼ X. Proof. Let f : Aω → Aω be a parity-reversing automorphism. Fix X and assume 2 ∼ ω ω A X = X. Writing X as A (I[u]2 ) and AX as A (J[u]2 ), we may define an isomorphism from X to AX by the map (u, x) 7→ (f(u), x). This map is well-defined: the interval (u, ·) in X is of type I[u]2 , and the interval (f(u), ·) in AX is of type ∼ J[au]2 = I[u]2 since f(u) ∈ [au]2. Hence X = AX. In the context of the cube problem, the case of interest is when A = X. If X is an order such that X3 =∼ X, then we may rewrite this as X2X =∼ X. By 3.13, if Xω has a p.r.a., then XX =∼ X as well, i.e. X2 =∼ X. In Section 5, parity- reversing isomorphisms for Aω are constructed for many different kinds of orders A, culminating in the proof that if X3 =∼ X then indeed Xω has a p.r.a., no matter the cardinality of X. 3.4. The isomorphism AnX =∼ X. We provide the analogous definitions and results (without proof) for analyzing orders that satisfy the isomorphism AnX =∼ X, for some fixed n > 1. Definition 3.14. Two sequences u, v ∈ Aω are called n-tail-equivalent if there exist finite sequences r, s ∈ A<ω with |r| ≡ |s| (mod n) and a sequence u0 ∈ Aω 0 0 such that u = ru and v = su . If u and v are n-tail-equivalent, we write u ∼n v.

It may be checked that ∼n is an equivalence relation. The ∼n-class of u is n ∼ denoted [u]n. The characterization of the isomorphism A X = X is given by the following theorem. Theorem 3.15. For an order X, we have AnX =∼ X if and only if X is isomorphic ω to an order of the form A (I[u]n ). The next theorem describes the relationship between the various orders AkX, 1 ≤ k ≤ n − 1, in the case where AnX =∼ X. The notation ak denotes the sequence aa . . . a ∈ A<ω in which a is repeated k times. Proposition 3.16. Suppose X is an order such that AnX =∼ X, so that X is of the ω k ∼ ω form A (I[u]n ). For a ﬁxed k, 1 ≤ k ≤ n − 1, let Y = A X. Then Y = A (J[u]n ), ω k where for all u ∈ A and a ∈ A, we have J[u]n = I[a u]n . k In the above, the sequence a may be replaced with any k-sequence a0a1 . . . ak−1 of elements of A. Deﬁnition 3.17. An order automorphism f : Aω → Aω is called an n-revolving ω automorphism (abbreviated n-r.a.) if for every u ∈ A and a ∈ A the image of [u]n under f is [au]n. Proposition 3.18. If AnX =∼ X and there exists an n-revolving automorphism on Aω, then AX =∼ X as well (and hence AkX =∼ X for all k). THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 15

Hence if Xn+1 =∼ X for some n ≥ 1 and there is an n-revolving automorphism on Xω, we have X2 =∼ X as well. 3.5. A solution to the cube problem when X is countable, without endpoints. As a corollary to Theorem 3.5, we prove the characterization that was claimed in Section 2 for countable orders X without endpoints such that Xn =∼ X for some n > 1. A consequence is that the cube property holds for such orders. The argument diﬀers substantially from the argument for the general case, and if desired this section can be skipped. Corollary 3.19. Let X be a countable linear order without endpoints. If Xn =∼ X for some n > 1, then X is isomorphic to an order of the form Q(Iq), where for each q, there are densely many p ∈ Q with Iq = Ip. Hence for any countable order L we have LX =∼ X, and in particular X2 =∼ X. Proof. Let A = Xn−1. The hypothesis is that AX =∼ X. By Theorem 3.5, X ω is isomorphic to an order of the form A (I[u]). Note that since A is countable, each equivalence class [u] is countable: each v ∈ [u] is of the form ru0 for some tail-sequence u0 of u, and there are only countably many r ∈ A<ω, and countably many tail-sequences u0. Since X is countable, it must be that for all but countably many of the equivalence classes [u] we have I[u] = ∅. Let k be the number of classes for which I[u] is nonempty, so that 1 ≤ k ≤ ω. Enumerate these classes as Ci, i < k, and let Ii S denote I[u] if [u] = Ci. Let C = i vn, and b, c ∈ A such that b < v0 and c > w0. Let x = (v n)au, y = bu, and z = cu. Then ω y < v < x < w < z in A , and clearly x, y, z ∈ [u] = Ci. But this proves the claims: Ci is dense in C, but further C is dense and without a top or bottom point. The proof is complete. Although we are going to show that the implication Xn =∼ X =⇒ X2 =∼ X holds for any linear order X, the advantage of the corollary is that it yields a complete classiﬁcation in the countable, no endpoints case. Such an X is of the form Q(Ii). Hence to specify X it is enough to specify the number of parts k in the partition S Q = i

When X is countable and has a single endpoint, the same proof goes through and yields an analogous classification. In the countable, two endpoints case, the proof does not carry over. The reason is that, while there is a unique (up to isomorphism) countable dense order with two endpoints, namely Q = Q ∩ [0, 1], this order does not enjoy the same invariance under multiplication that Q enjoys. There exist countable orders L with both endpoints such that LQ =6∼ Q. In fact it is easy to see we have non-isomorphism whenever L is not dense. Moreover, this difference has consequences. Unlikes for Q, there are right products of Q that are not isomorphic to their squares. For example, X = Q2 is not isomorphic to its square. It follows that the analogous classification in this case fails, though it may be that some more detailed classification is possible. In the uncountable case, the proof breaks down completely, and the results of Section 5 show that no similar classification is possible. In many models of set theory, there do not even exist analogies to the order Q on higher cardinals. Even n ∼ ∼ when such orders do exist, nothing like “X = X if and only if X = Q(Ii)” is true for uncountable X. For example, under the continuum hypothesis there exists the saturated order Q1 of size ℵ1. This order has a characterization a la Cantor’s characterization of Q, and enjoys many analogous properties to those of Q. For one, if L is any order of ∼ 2 ∼ size at most ℵ1, then LQ1 = Q1, and in particular Q1 = Q1. Furthermore, we have the result “any order X of the form Q1(Ii) is invariant under left multiplication by any L of size at most ℵ1. In particular, if the orders Ii are of size at most ℵ1, then n ∼ for any n we have X = X.” (Here, each Ii replaces densely many points in Q1.) n ∼ But the converse is false outright: if X is of size ℵ1 and X = X it need not be ∼ true that X = Q1(Ii). For example, it is possible to use the methods of Section 5 to 2 ∼ produce, without extra set theoretic hypotheses, an X of size ℵ1 such that X = X ∼ but ω1X =6 X.

4. Parity-reversing automorphisms on Aω In this section we show how to construct parity-reversing automorphisms of Aω for orders A satisfying certain structural requirements. A consequence of our work is that if X3 =∼ X, then Xω has a parity-reversing automorphism. By Proposition 3.13 it follows that X2 =∼ X as well. At the end of the section, we sketch how to generalize the argument to get the implication Xn =∼ X =⇒ X2 =∼ X for any order X and n ≥ 2, finishing the proof of the main theorem. ω For a sequence u = (u0, u1, u2,...) ∈ A , let σu denote the once shifted tail- n sequence (u1, u2,...) and σ u the n-times shifted tail-sequence (un, un+1,...). As a first step toward constructing a p.r.a. for Aω, we show that any closed interval in Aω with periodic endpoints of odd period has a parity-reversing automorphism. It is then possible to write down conditions under which a collection of these partial maps can be stitched together to get a full p.r.a. on Aω. A technical note: if r, s ∈ A<ω are finite sequences, it can be shown that if r < s in Aω, then it always holds that r < rs < sr < s, even when one of the sequences is an initial sequence of the other. We will use this fact repeatedly. However, in every case in the subsequent work where we actually consider specific sequences r < s, it will be apparent that we actually have r < rs < sr < s. Now, suppose that we have finite sequences r, s ∈ A<ω of lengths m, n respectively, such that r < s in Aω. The shift map u 7→ σmu restricted to the interval THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 17

[r, rs] is an isomorphism of this interval with the interval [r, sr]. The inverse map is given by u 7→ ru, restricted to [r, sr]. Similarly the map u 7→ su, when restricted to [rs, s], is an isomorphism of this interval with [sr, s] whose inverse is given by u 7→ σnu. Thus we have the following proposition. Lemma 4.1. Suppose r, s ∈ A<ω are finite sequences, respectively of lengths m, n, such that r < s in Aω. Then the map f :[r, s] → [r, s] defined by σmu u ∈ [r, rs] f(u) = su u ∈ [rs, s] is an order automorphism of the interval [r, s]. Its inverse is given by ru u ∈ [r, sr] f −1(u) = σnu u ∈ [sr, s]. Note that there is no ambiguity in the definition of f(rs) in the statement of the lemma, since σmrs = srs = sr. Given an interval of the form [r, s] in Aω, we call the automorphism f defined in Lemma 4.1 a standard map on [r, s]. Standard maps are the essential tool in our proof of the cube property for (LO, ×). Standard maps are defined with respect to given sequences r, s ∈ A<ω, and an interval can carry more than one standard map. For example, if r0 = a, r1 = aa, and s = b for some points a < b in A, then we have [r0, s] = [r1, s] = [a, b]. However, the standard map on this interval defined with respect to the sequences r0, s is different than the standard map defined with respect to the sequences r1, s. We adopt the convention that the phrase “the standard map on [r, s]” means the one defined with respect to the sequences r, s. Lemma 4.2. Fix r, s ∈ A<ω of lengths m, n respectively. If m, n are both odd, then the standard map f :[r, s] → [r, s] is parity-reversing, in the sense that for all u ∈ [r, s] and a ∈ A, we have f(u) ∈ [au]2. Proof. Either f(u) = σmu or f(u) = su. In either case, the obvious meeting representation between au and f(u) witnesses f(u) ∼2 au. Theorem 4.3. Suppose A has both a left and right endpoint. Then Aω has a parity-reversing automorphism. Proof. Let 0 denote the left endpoint of A, and 1 denote the right endpoint. Then Aω also has left and right endpoints, namely 0 and 1. That is, Aω = [0, 1]. Thus the standard map on this interval is actually an automorphism of Aω. It is parity- reversing by Lemma 4.2, since 0 and 1 are both of period 1. Thus if A has both a left and right endpoint, and X is an order such that 2 ∼ ∼ ω A X = X, then AX = X as well. Decomposing X as A (I[u]2 ) and AX as ω A (J[u]2 ), the isomorphism is given by (u, x) 7→ (f(u), x), where f is the standard map on Aω = [0, 1]. In this case, we may alternatively apply Lindenbaum’s version of the Schroeder- Bernstein theorem to get an isomorphism between X and AX: X, which is isomorphic to A2X, contains an initial copy of AX by virtue of the fact that A has a left endpoint, and AX contains a final copy of X since A has a right endpoint. Thus AX =∼ X by Lindenbaum’s theorem. If A = X, we recover Corollary 2.1. 18 GARRETT ERVIN

These approaches are actually the same. The isomorphism one gets from the proof of Lindenbaum’s theorem (which is really just the classical proof of the ω Schroeder-Bernstein theorem), when viewed as an isomorphism of A (I[u]2 ) with ω A (J[u]2 ), turns out to be exactly the isomorphism (u, x) → (f(u), x), where f is the standard map on [0, 1] = Aω. In this sense, standard maps are generalized instances of the classical Schroeder-Bernstein bijection in the context of orders of the form Aω. The standard map f fixes the endpoints 0 and 1 of Aω—necessarily so, since f is an order automorphism. It is easily checked that these are the only fixed points of f. This does not change the fact that f is parity-reversing: both 0 and 1 are periodic sequences of period 1, so that [a0]2 = [0]2 = [0] and [a1]2 = [1]2 = [1] for any a ∈ A. Thus the parity-reversing requirement on f at these points is simply that f(0) ∈ [0] and f(1) ∈ [1], which certainly holds. Viewing f as defining an ∼ 2 ∼ ω ∼ ω isomorphism between some X = A X = A (I[u]2 ) and AX = A (J[u]2 ), we have that this isomorphism maps I0 onto J0, which is legitimate as these intervals are identical, as noted in the discussion following 3.11. Similarly for I1 and J1. If A does not have any endpoints, or only a single endpoint, then Aω is not of the form [r, s]. Thus no standard map defines a p.r.a. on all of Aω. However, because they fix the endpoints of the intervals on which they are defined, standard maps can be stitched together to obtain automorphisms of longer intervals. For example, suppose r, s, t ∈ A<ω are sequences such that r < s < t. Let f denote the standard map on [r, s] and g the standard map on [s, t]. Then f ∪ g is an automorphism of [r, t]. Its fixed points are exactly r, s, and t. If |r|, |s|, and |t| are all odd, then f and g, and thus f ∪ g, are parity-reversing. The standard map h on [r, t] also serves as an automorphism of this interval. This map is different than the two-piece map f ∪ g (e.g. h does not fix s). The advantage of the piecewise construction is that it can be extended to get parity- reversing maps on intervals without endpoints, as well as orders without endpoints. To deal with such intervals, we introduce some terminology. The cofinality of an interval I is the minimum length λ of an increasing sequence of points {xi : i < λ} ⊆ I such that for every y ∈ I there is i < λ with y ≤ xi. Similarly, the coinitiality of I is minimum length κ of a decreasing sequence in I that eventually goes below every element of I. The cofinality is 1 if the interval has a maximal element, and if it has a minimal element the coinitiality is 1. When these cardinals are not 1, they are infinite and regular. An interval with coinitiality κ and cofinality λ is called a (κ, λ)-interval. We usually write these as ordinals, e.g. ω, ω1, etc., to emphasize that they refer to sequences. We refer to (1, 1)-intervals as 2-intervals, (1, ω)-intervals as ω-intervals, (ω, 1)-intervals as ω∗-intervals, and (ω, ω)-intervals as Z-intervals, since these intervals are, respectively, spanned by sequences of order type 2, ω, ω∗, and Z. Viewing the order A as itself an interval, we may speak of (κ, λ)-orders, or 2, ω, ω∗, and Z-orders. A cover for an order A is a collection of disjoint intervals C = {Ca : a ∈ A} so that a ∈ Ca for all a. Indexing by the elements of A is for convenience: it is not assumed that a 6= b implies Ca 6= Cb, since this would give only trivial covers, but only that either Ca = Cb or Ca ∩ Cb = ∅. A cover C is called a Z-cover if every C ∈ C is a Z-interval. Not every order A admits a Z-cover: if, for example, A has a left endpoint 0, then in any cover C for A, the interval C0 cannot be a Z-interval. Less trivially, suppose A is a complete (ω1, THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 19

ω1)-order, and C is a cover for A. Suppose C contains a Z-interval C. This interval is open. Since A is “long” to both the left and right, C is bounded. Since A is complete, C has a greatest lower bound x and least upper bound y. The intervals Cx and Cy must be disjoint from C, hence x must be the maximal element in Cx and y the minimal element in Cy. Thus neither is a Z-interval, and it follows that A admits no Z-cover. A cover C is called a {Z, ω}-cover if every C ∈ C is either a Z-interval or an ω-interval. Similarly, C is called a {Z, ω∗}-cover if every C ∈ C is either a Z-interval ∗ or ω -interval. Given a cover C, let CX ⊆ C denote the collection of intervals in C of type X. For example, if C is a {Z, ω}-cover, then C = CZ ∪ Cω. Theorem 4.4. 1. If A admits a Z-cover, then Aω has a parity-reversing automorphism. 2. If A has a left endpoint and admits a {Z, ω}-cover, then Aω has a parity- reversing automorphism. 3. If A has a right endpoint and admits a {Z, ω∗}-cover, then Aω has a parity- reversing automorphism. Thus in any of these three cases, if X is an order such that A2X =∼ X, then AX =∼ X as well. Proof. (1.) Assume first that A has a Z-cover C. It follows that A has no endpoints. C C C C For every C ∈ C, fix a Z-sequence . . . < x−1 < x0 < x1 < x2 < . . . spanning C. C C To each of the points xk in A there is the corresponding periodic sequence xk in ω C C D D A . For C,D ∈ C and k, l ∈ Z, the intervals [xk , xk+1) and [xl , xl+1) intersect if and only if they are identical. The same is true of the corresponding intervals C C D D ω [xk , xk+1) and [xl , xl+1) in A . These intervals actually cover Aω, that is, for every u ∈ Aω there is a unique C C C ∈ C and k ∈ Z such that u ∈ [xk , xk+1). To see this, note that since the intervals C cover A, u’s first entry u0 (viewed as a point in A) falls in one of the C. There C C C are two possibilities: either xk < u0 < xk+1 for some k ∈ Z, or u0 = xk for some C C C C k ∈ Z. In the first case, we have that xk < u < xk+1 so that u ∈ [xk , xk+1). In the C C C C second, either xk−1 < u < xk or xk ≤ u < xk+1 depending on whether the first C C entry of u differing from xk (if it exists) is greater or less than xk . Thus either C C C C u ∈ [xk−1, xk ) or u ∈ [xk , xk+1). We have shown that ω [ C C A = [xk , xk+1]. k∈Z C∈C The intervals in this union are pairwise disjoint unless they are consecutive, in which case they share a single endpoint. C C C For every k ∈ Z and C ∈ C, let fk denote the standard map on [xk , xk+1]. Since C the endpoints of this interval are of period 1, we have that fk is parity-reversing. The function [ C f = fk k∈Z C∈C C is well-defined, since the domains of two different fk share at most one point and at C that point the functions agree. Since each fk is a parity-reversing automorphism C C ω ω of [xk , xk+1] and A is the union of these intervals, f is a p.r.a. for A . 20 GARRETT ERVIN

(2.) Now assume that A has a left endpoint 0 and a {Z, ω}-cover C. Then A has no right endpoint. The argument is similar to the previous one, but with a catch. C C C C For every C ∈ CZ, fix a sequence . . . < x−1 < x0 < x1 < x2 < . . . spanning C. C C C Similarly, for every C ∈ Cω fix a sequence x0 < x1 < x2 < . . . spanning C, where C x0 is the left endpoint of C. C C As before, the intervals [xk , xk+1) are pairwise disjoint. The difference is that now there may be points in Aω that do not fall in any of these intervals. To see this, suppose u ∈ Aω. Then there is a unique C ∈ C and k such that C C C C C C xk ≤ u0 < xk+1. If in fact xk < u0 < xk+1, then certainly u ∈ [xk , xk+1). So C suppose u0 = xk . If either C ∈ CZ, or C ∈ Cω and k > 0, then u is contained in C C C C either [xk−1, xk ) or [xk , xk+1), depending on whether the first entry of u differing C C from xk (if it exists) is greater than or less than xk . C C C So suppose u0 = x0 for some C ∈ Cω. Then either u ≥ x0 or u < x0 . In the C C first case we have u ∈ [x0 , x1 ). C The issue occurs in the second case when u < x0 . Assume we are in this case, C and for notational simplicity denote x0 by x. It must be then, that x (viewed as a point in A) is greater than the left endpoint 0, and further that un < x, where un is the leftmost entry of u differing from x. However, since u0 = x it must be that u ≥ x000 ... = x0. Thus u ∈ [x0, x). This interval is disjoint from all of the C C ω [xk , xk+1), and by the same argument, any v ∈ A that is not contained in one of C C the [xk , xk+1) must be contained in an interval of this form. Thus

ω [ C C [ C C [ A = [xk , xk+1] ∪ [xk , xk+1] ∪ [x0, x]. k∈ k∈ω C Z x=x0 C∈C C∈Cω Z C∈Cω

C C These intervals are pairwise disjoint up to endpoints. The intervals [xk , xk+1] have C parity-reversing standard maps fk . If we can show that for every interval [x0, x], there is a parity-reversing automorphism fx :[x0, x] → [x0, x], then the map

[ C [ C [ f = fk ∪ fk ∪ fx k∈ k∈ω C Z x=x0 C∈C C∈Cω Z C∈Cω is a parity-reversing automorphism of Aω. C So fix C ∈ Cω and let x = x0 . If C is the leftmost interval in C, then x = 0 and ω [x0, x] = {0} is just the left endpoint of A . In this case let fx be the map that fixes 0 and is undefined elsewhere. Otherwise x > 0. The interval [x0, x] has a periodic right endpoint, but only an eventually periodic left endpoint. We have not defined a standard map for such an interval. Note however that the shift map u 7→ σu restricted to [x0, x] is an isomorphism of this interval with [0, x]. Let g denote this shift map, and let f −1 denote the standard map on [0, x]. Then fx = g ◦ f ◦ g is a parity-reversing automorphism of [x0, x], as desired. Now f, as defined above, is a p.r.a. for Aω.

(3.) Finally, suppose A has a right endpoint and admits a {Z, ω∗}-cover. Then A∗ has a left endpoint and admits a {Z, ω}-cover. Thus (A∗)ω, which is isomorphic THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 21 to (Aω)∗, has a p.r.a. f ∗. The corresponding automorphism f on Aω is still parity- reversing since the requirement f(u) ∈ [au]2 does not depend on the ordering of ω A , but only on the underlying set of points. Suppose for the moment that A is an order without endpoints. The content of the theorem is that if A has a Z-cover, then Aω can be covered disjointly (up to endpoints) by intervals of the form [x, y], with x, y ∈ A. Since we have parity- reversing standard maps on these intervals, we can use this decomposition to build a p.r.a. on all of Aω. The essence of what can go wrong when A does not admit a Z-cover can be illustrated by an example. Suppose that A is a complete (ω1, ω1)-order. We have seen that A has no Z-cover. We might still attempt to build a p.r.a. for Aω piecewise: begin with some x0 < x1 in A and consider the corresponding periodic ω points x0 < x1 in A . Put the standard map on [x0, x1]. Then pick some x2 > x1, and so on. After ω-many steps we will have defined a p.r.a. on the interval spanned by x0 < x1 < x2 < . . . . Now, since A is complete and of cofinality ω1, the sequence x0 < x1 < x2 < . . . is bounded in A and converges to some point x. However, the sequence of xn does not converge to x in Aω. Rather, there is a nonempty open interval I sitting above all the xn and below x. This interval I consists of points u with first coordinate u0 = x but with some later coordinate ui < x. Notice that there are no periodic points of period 1 in this interval. We might hope to bridge this gap (perhaps up to some collection of legally fixable points) with intervals whose endpoints are only eventually periodic, as we did in the proof of (2.) of Theorem 4.4. There, however, it was crucial for the argument that A had a left endpoint 0. It turns out it is impossible to bridge this gap, in the sense that any automorphism on Aω extending the partial automorphism described above cannot be parity-reversing on I. Moreover, no alternative construction works. We will show in Section 6 that any complete (ω1, ω1)-order does not admit a parity-reversing automorphism. Here is an immediate corollary of Theorem 4.4.

Corollary 4.5. If A is a 2-order, ω-order, ω∗-order, or Z-order, then Aω has a parity-reversing automorphism. Thus for any order X, if A2X =∼ X then AX =∼ X. Proof. When A is a 2-order, this is simply a restatement of Theorem 4.3. In the other three cases A has, respectively, a {Z, ω}-cover, {Z, ω∗}-cover, or Z-cover given by C = {A}. Among the orders satisfying the hypotheses of the corollary are all countable orders, since any countable order has a coﬁnality and coinitiality at most ω. Thus for any countable A and any X we have A2X =∼ X =⇒ AX =∼ X. This gives in particular that any countable X isomorphic to its cube is isomorphic to its square, recovering Corollaries 2.1 and 3.19 and also giving the result for countable orders with a single endpoint. The order R of the real numbers gives an example of an uncountable Z-order. Hence R2X =∼ X =⇒ RX =∼ X for any X. One of the simplest examples of an order that is not a Z-order but admits a Z-cover is ω1Z. Here, the cover consists of the intervals Cα = {(α, z): z ∈ Z}. Similarly, ω1R admits a Z-cover, since each copy of R is spanned by a Z-sequence. 22 GARRETT ERVIN

The order ω1 has a left endpoint, but admits no {Z, ω}-cover, as can be shown by a similar argument to the one showing any complete (ω1, ω1)-order has no Z-cover. ω Thus it does not follow from 4.4 that there is a p.r.a. on ω1 . It turns out that ω ∗ ω1 does have a p.r.a., though we will not prove this. However, if A = ω1 + ω1 is ∗ the order obtained by putting a copy of ω1 to the right of a copy of ω1 , then A is a complete (ω1, ω1)-order. It follows from the results of Section 6 that there is no p.r.a. on Aω. We will show in fact that there exists an X such that A2X =∼ X but AX =6∼ X. It remains to show that any order X that is isomorphic to its cube admits the right kind of cover to get a p.r.a. on Xω. This is Theorem 4.6 below. When combined with Theorem 4.4, it shows that for any X without endpoints or with a single endpoint, if X3 =∼ X then X2 =∼ X. Since we have already dealt with the two endpoint case, we have X3 =∼ X =⇒ X2 =∼ X for any X, finishing the proof of the cube property. Theorem 4.6. Suppose X is an order such that X3 =∼ X. 1. If X has no endpoints, then X admits a Z-cover. 2. If X has a left endpoint but no right endpoint, then X admits a {Z, ω}- cover. 3. If X has a right endpoint but no left endpoint, then X admits a {Z, ω∗}- cover. The proof of the theorem is in two lemmas. Before we can state these lemmas, we need some more terminology. Up to this point we have only studied invariance of orders under left multiplication (by a fixed A or power of A). Orders X such that Xn =∼ X fit into this context since the isomorphism Xn =∼ X can be rewritten as An−1X =∼ X, where A = X. However, such orders also display an invariance under right multiplication, since Xn =∼ X can just as well be rewritten XAn−1 =∼ X, where again A = X. It is this right-sided invariance under multiplication that is needed to get the covers of X in Theorem 4.6. Forgetting the isomorphism X3 =∼ X for now, we consider the single power version of this right-sided invariance. That is, for an order A, we analyze the structure of orders X such that XA =∼ X. In the case of invariance under left multiplication by A, the space Aω of right- infinite sequences plays a crucial role. In the case of right-sided invariance, it is the space of left-infinite sequences that is relevant. ω∗ Let A = {(. . . , a2, a1, a0): ai ∈ A, i ∈ ω} denote the set of left-infinite sequences of elements of A. Notice that we still index the entries of such sequences by elements in ω. We copy over the notation from the right-sided case, reversing ∗ it when necessary. The letters u, v, . . . will now be used to denote elements of Aω . The nth entry of u is still denoted un. <ω∗ Let A = {(an−1, . . . , a1, a0): ai ∈ A, n ∈ ω} denote the set of left-growing ∗ finite sequences. The letters r, s, . . . will for now denote elements of A<ω . For <ω∗ ω∗ r = (an−1, . . . , a1, a0) ∈ A and u = (. . . , u1, u0) ∈ A , we use ur to denote the sequence (. . . , u1, u0, an−1, . . . , a1, a0). ∗ It is impossible to order Aω lexicographically, in the sense that two left-infinite sequences may not have a leftmost place in which they differ. If, however, two such sequences eventually agree, it is possible to compare them lexicographically. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 23

∗ Deﬁnition 4.7. For u, v ∈ Aω , we say u is eventually equal to v, and write u ∼∞ v, if there exists N ∈ ω such that for all n > N we have un = vn.

Equivalently, u ∼∞ v if there exist finite sequences r, s with |r| = |s| and a ∗ sequence u0 ∈ Aω such that u = u0r and v = v0s. This is just the usual eventual equality relation, considered here for left-infinite sequences. The ∼∞-class of u is denoted [u]∞. ω∗ Observe that for a given u ∈ A , the class [u]∞ can be ordered lexicographically. If u ∼∞ v, define u < v if and only if uN < vN , where N is the leftmost place such that uN 6= vN . It is immediate that u 6< u, and one of u < v, u = v, u > v always holds. Transitivity can be checked as well. ω∗ The classes [u]∞ are the largest subsets of A that can possibly be ordered lexicographically, in the sense that if u 6∼∞ v, then u and v have no leftmost place of difference. In what follows, whenever we refer to an order on [u]∞ we mean the lexicographical order. If we write u < v it is assumed that u ∼∞ v. Here are our lemmas.

∗ Lemma 4.8. Fix u ∈ Aω .

1. If A has no endpoints, the class [u]∞ is a Z-order. 2. If A has a left endpoint but no right endpoint, the class [u]∞ is either an ω-order or Z-order. 3. If A has a right endpoint but no left endpoint, the class [u]∞ is either an ω∗-order or Z-order. Proof. (1.) Assume first A has no endpoints. We define a sequence . . . < v−1 < 0 1 0 v < v < . . . spanning [u]∞. Let v = u. For n a fixed positive integer, define n n n n n v = (. . . , v1 , v0 ) to be any sequence such that vm = um for all m > n but vn > un. n It is always possible to find such a vn, since A does not have a top point. On the other side, again for n a fixed positive integer, let v−n be a sequence such that −n −n vm = um for all m > n, but now vn < un. This is possible since A does not have a bottom point. It is clear that . . . < v−1 < v0 < v1 < . . . since the ordering is lexicographical. Further if v ∈ [u]∞, then there is some N such that for all m ≥ N we have vm = um. −N N n Thus v < v < v , and so the sequence v spans [u]∞, as desired.

(2.) Now assume A has a left endpoint 0 but no right endpoint. The class [0]∞ has a left endpoint, namely 0 itself. Let v0 = 0 and vn be any sequence that is 0 beyond the nth place, but is greater than 0 in the nth place. Then we have 0 1 v < v < . . . as before, and in fact this sequence spans [0]∞. Thus [0]∞ is an ω-order. So assume u 6∼∞ 0. We show [u]∞ is a Z-order. As before, since A has no top 1 2 point, we may find v < v < . . . cofinal in [u]∞. On the other side, observe that since u 6∼∞ 0, there are infinitely many places n n such that un > 0. Enumerate these places as nk, k ∈ ω. Let v k be any sequence that agrees with u beyond the n th place, but such that vnk < u . This is possible k nk nk n1 n0 since unk > 0. Then . . . < v < v , and further this sequence is coinitial in [u]∞. Hence [u]∞ is a Z-order, as claimed. The case (3.) is symmetric to (2.). Lemma 4.9. Suppose that X is an order such that XA =∼ X. 1. If A has neither a left nor right endpoint, then X admits a Z-cover. 24 GARRETT ERVIN

2. If A has a left endpoint but no right endpoint, then X admits a {Z, ω}- cover. 3. If A has a right endpoint but no left endpoint, then X admits a {Z, ω∗}- cover. Proof. The intuition of the proof is simple. If XA =∼ X, then X can be organized into X-many intervals of type A. Since there are X-many of them, these intervals in turn may be organized into XA-many copies of A, that is, X-many copies of A2. And so on. Now consider the situation from the point of view of some fixed x ∈ X. At the first stage x is included in a copy of A, and in the second, in some copy of A2 containing the initial copy of A, etc. These larger and larger intervals surrounding x, consecutively isomorphic to A, A2, A3, ..., have a limit, which is ω∗ isomorphic to [u]∞ for some u ∈ A . The conclusion follows. To make this explicit, let f : X → XA be an isomorphism. Let fl and fr denote the left and right components of f, that is, the unique functions such that f(x) = (fl(x), fr(x)) for all x ∈ X. For every x ∈ X, we define a sequence of points x x x0, x1,... in X and a sequence a0 , a1 ,... in A. Let x0 = x and recursively define x xn+1 = fl(xn). Let an = fr(xn). x With this notation, we have f(x) = (x1, a0 ). By repeatedly factoring the xi, n x x x we get an isomorphism from X onto XA defined by x 7→ (xn, an−1, . . . a1 , a0 ). Although it is not literally an n-fold composition of f, we denote this isomorphism by f n. x For n ∈ ω, define In to be the set of y ∈ X such that yn = xn. Then the image x n n of In under f is the set of points in XA of the form (xn, bn−1, . . . , b0), bi ∈ A. x n Hence In is an interval in X, and it is isomorphic to A . Furthermore, we have the x x x containments {x} = I0 ⊆ I1 ⊆ I2 ⊆ ... since if yN = xN then for every n ≥ N we have yn = xn as well. x S x x x Let I∞ = n In. Then since the In form a chain of intervals in X, I∞ is also an interval in X. Notice that by definition, for every N ∈ ω and x, y ∈ X, either x y x y x y x IN ∩ IN = ∅ or IN = IN , and in this latter case In = In for all n ≥ N. Hence I∞ y and I∞ are either equal or disjoint as well. x x x For x ∈ X, define ux to be the sequence (. . . , a1 , a0 ). Then if y ∈ I∞, it must be that for all sufficiently large n we have yn = xn. Thus for all sufficiently large x y n we have an = an, which gives ux ∼∞ uy. On the other hand, suppose v ∈ [ux]∞, x x say v = (. . . , an+1, an, bn−1, . . . , b1, b0) for some bi ∈ A, i < n. There is a unique n y y y ∈ X such that f (y) = (yn, an−1, . . . , a0) = (xn, bn−1, . . . , b0). But then uy = v. x This shows that the map F : I∞ → [ux]∞ defined by F (y) = uy is a bijection. x This map is clearly order-preserving as well, and hence an isomorphism of I∞ with x [ux]∞. Thus {I∞ : x ∈ X} is a cover of X by intervals of the form [u]∞. The conclusion of the lemma now follows from Lemma 4.8. Proof of Theorem 4.6. Suppose X is an order such that X3 =∼ X. Then XA =∼ X, where A = X2. If X has no endpoints, then A has no endpoints. By Lemma 4.9, X admits a Z-cover, and therefore, by Lemma 4.8, Xω admits a parity-reversing automorphism. If X has a left endpoint, but not a right one, then similarly A has a left endpoint, but no right one. Hence X admits a {Z, ω}-cover, and again Xω admits a parity-reversing automorphism. The right endpoint case is symmetric. We conclude this section with a sketch of the proof that Xn =∼ X implies X2 =∼ X for all orders X and all n ≥ 2. When n = 2 the statement is trivial, and we have THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 25 just finished the proof for n = 3. The argument for larger n is a straightforward adaptation of the case when n = 3. For convenience, we consider orders satisfying the (only notationally distinct) isomorphism Xn+1 =∼ X, and we assume n > 2. This isomorphism can be rewritten as AnX =∼ X, where A = X. We know by the results at the end of Section 3 that if Aω has an n-revolving automorphism (n-r.a.), then AX =∼ X. So we turn to the question of when Aω admits such an automorphism. We build n-r.a.’s as we built p.r.a.’s, that is, as unions of standard maps. In our previous argument, to ensure that a map f was parity-reversing, it was enough to have that for every u either f(u) = σnu or f(u) = ru, with n and |r| both odd. This is because deleting or adding an initial sequence of odd length always sends u into [au]2. For n > 2, the situation is not symmetric. Lemma 4.10. Suppose f : Aω → Aω is an order automorphism, and for every u ∈ Aω, either f(u) = σku for some k ≡ n − 1 (mod n) or there exists a finite sequence r with |r| ≡ 1 (mod n) and f(u) = ru. Then f is an n-r.a.

Proof. In either case, the obvious meeting representation witnesses f(u) ∼n au. Thus if we have two sequences r, s ∈ A<ω such that |r| ≡ n − 1 (mod n) and |s| ≡ 1 (mod n), the standard map on [r, s] is n-revolving. As before, when A has both a left and right endpoint, a single standard map serves to get an n-r.a. of Aω. Theorem 4.11. Suppose A has both a left and right endpoint. Then Aω has an n-revolving automorphism. Proof. Let 0 and 1 denote the left and right endpoints of A. Then Aω = [0n−1, 1]. By Lemma 4.10, the standard map on this interval is n-revolving. In particular, if X has both endpoints and Xn+1 =∼ X, then X2 =∼ X. For the cases with one or neither endpoint, we need the generalizations of Theorems 4.4 and 4.6. Theorem 4.12. 1. If A admits a Z-cover, then Aω has an n-r.a. 2. If A has a left endpoint and admits a {Z, ω}-cover, then Aω has an n-r.a. 3. If A has a right endpoint and admits a {Z, ω∗}-cover, then Aω has an n-r.a. Thus in any of these three cases, if X is an order such that AnX =∼ X, then AX =∼ X as well. Theorem 4.13. Suppose X is an order such that Xn+1 =∼ X. 1. If X has neither a left nor right endpoint, then X admits a Z-cover. 2. If X has a left endpoint but no right endpoint, then X admits a {Z, ω}- cover. 3. If X has a right endpoint but no left endpoint, then X admits a {Z, ω∗}- cover. The conjunction of these theorems along with 4.11 gives that Xn+1 =∼ X =⇒ X2 =∼ X for all X. Theorem 4.13 follows immediately from Lemmas 4.8 and 4.9. The proof of 4.12 (1.) is essentially the same as 4.4 (1.): if A has a Z-cover, then Aω can be covered (disjointly up to endpoints) by intervals of the form [a, b] for a, b ∈ A. Writing such intervals as [an−1, b], we have that the associated standard maps are n-revolving. The union of these maps then gives an n-r.a. on Aω. 26 GARRETT ERVIN

The proof of 4.12 (2.) is also very similar to that of 4.4 (2.). Given a {Z, ω}-cover of A we obtain a cover of Aω by intervals of the form [a, b] and [xa, b]. Intervals of the ﬁrst type have n-revolving standard maps, as above. Intervals of the second type shift onto intervals of the ﬁrst type, and therefore also have n-revolving automorphisms. The union of these maps yields an n-r.a. for Aω. The proof of (3.) is symmetric. The main theorem follows:

Theorem 4.14. Suppose X is an order such that Xn =∼ X for some n > 1. Then X2 =∼ X. In particular, the cube property holds for (LO, ×).

5. Constructing orders X such that Xn =∼ X The purpose of this section is to justify the previous work, in two ways. First, we will show that the cube problem for linear orders, and more generally the problem of showing Xn =∼ X =⇒ X2 =∼ X, is not vacuous, in the sense that for every n there exist many orders X such that Xn =∼ X. We construct such orders below as direct limits, and show in particular that they can be of any infinite cardinality. Secondly, we wish the justify the need for the machinery developed in Section 4 for solving the cube problem. When we say “X3 =∼ X,” what is meant implicitly is that there exists an isomorphism f : X3 → X. All of our analysis of the relation X3 =∼ X has really been with respect to a fixed isomorphism f. Associated to such ω ω an f is an order of the form X (I[u]2 ) and an isomorphism F : X → X (I[u]2 ) built 2 ω using f. If we view f, via the relabeling F , as an isomorphism of X × X (I[u]2 ) ω with X (I[u]2 ), then f is just the flattening isomorphism fl2 on the first three coordinates. This follows from the proof of 3.5 and the discussion afterwards. Conversely, if we can ever construct an isomorphism F from X onto an order of the ω 3 ∼ form X (I[u]2 ), we immediately have that X = X as witnessed by the flattening isomorphism. Suppose it were the case that for every X for which there exists an isomorphism ω ∼ F of X onto an order of the form X (I[u]2 ), we had that I[u]2 = I[au]2 for every ω a ∈ X and u ∈ X . Let us call such a decomposition trivial. Then letting I[u] ∼ ω denote the common order type of I[u]2 and I[au]2 , we would have that X = X (I[u]) and hence X2 =∼ X. We would have this isomorphism of X with X2 without any need for a parity-reversing automorphism of Xω, and the work in Section 4 showing that Xω has such an automorphism would be unnecessary to solve the cube problem. However, we shall show that this is not the case. There are pairs (X,F ) where X is a linear order and F is an isomorphism of X onto an order of ω ∼ the form X (I[u]2 ) such that for many a and u we have I[au]2 =6 I[u]2 . This follows from Theorem 5.2 below. For such an X we have X3 =∼ X naturally, but in order to show X2 =∼ X we need the device of a p.r.a. for Xω. Although Theorem 5.1 can be derived from 5.2, it has a proof which is easier to understand and thereby serves as a warmup for the proof of 5.2. There are proofs of these theorems in more algebraic language, but we elect to keep the presentation elementary. Given orders X and Y , an embedding of X into Y is an injective order-preserving map f : X → Y . Given embeddings f : X0 → Y0 and g : X1 → Y1 the map defined by (x, y) 7→ (f(x), g(y)) is an embedding of X0 × X1 into Y0 × Y1. We denote this map by (f, g). THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 27

S Given a sequence of orders X0 ⊆ X1 ⊆ X2 ⊆ ... the order X = i∈ω Xi is well-deﬁned. Similarly, given a directed system

f0 f1 X0 −→ X1 −→ ... where each fi is an embedding of Xi into Xi+1, we may form the direct limit X = lim X . −→ i i

If the fi are inclusion maps, the direct limit of the Xi is isomorphic to their union. We will sometimes confuse the direct limit construction with the union construction, and speak of each X as a suborder of X for j ≥ i, and as a suborder of X = lim X . i j −→ i Given two systems f0 f1 X0 −→ X1 −→ ... g0 g1 Y0 −→ Y1 −→ ... we obtain a system

(f0,g0) (f1,g1) X0 × Y0 −−−−→ X1 × Y1 −−−−→ ... It is a standard fact that if Z = lim(X × Y ), then Z ∼ X × Y , where X = lim X −→ i i = −→ i and lim Y . Written shortly, we have −→ i lim(X × Y ) ∼ (lim X ) × (lim Y ). −→ i i = −→ i −→ i Given orders X and Y , we say that X spans Y if there is an embedding f of X into Y such that for every y ∈ Y , there exist y0, y1 in the image f[X] such that y0 ≤ y ≤ y1. The following theorem says that any order can be expanded to an order of the same cardinality that is isomorphic to its square.

Theorem 5.1. Let X0 be any order. Then there exists an order X such that 1. X2 =∼ X, 2. X0 spans X, 3. |X| = |X0| + ℵ0. 2n Proof. For every i ∈ ω, let Xi+1 = Xi × Xi. Then Xn = X0 . Let f0 : X0 → X1 be the embedding deﬁned by f0(x) = (x, x). For every i > 0, let fi = (fi−1, fi−1). Then fi is an embedding of Xi into Xi+1. Let X = lim X . Then we have −→ i X = lim X (i ≥ 0) −→ i ∼ lim X (i > 0) = −→ i = lim(X × X )(i > 0) −→ i−1 i−1 = lim(X × X )(i ≥ 0) −→ i i ∼ lim(X ) × lim(X )(i ≥ 0) = −→ i −→ i = X × X. This proves (1.). The cardinality claim (3.) is clear. To verify (2.), let us view the fi as inclusions, so that each Xi is included in Xj for j ≥ i, and in X. Then X0 is included in X1 = X0 × X0 as the set of points of the form (a, a), and more 2n generally in Xn = X0 as the set of points of the form (a, a, . . . , a). Fix x ∈ X. Then x ∈ Xn for some n. Clearly, for some a, b we have (a, a, . . . , a) ≤ x ≤ (b, b, . . . , b). That is, x lies between two points of X0, provided we view X0 as a subset of Xn. Hence we also have that x lies between two points of X0, now viewing X0 as a subset of X. 28 GARRETT ERVIN

Note in particular that X has the same endpoint configuration as X0: if X0 has both endpoints, then so does X, and likewise for the other cases. Hence there are orders of any cardinality and any endpoint configuration isomorphic to their squares. We can get “X3 =∼ X” instead of “X2 =∼ X” in the conclusion of the theorem 3 by letting Xi+1 = Xi and letting f0 be the embedding x 7→ (x, x, x). Similarly, we can get orders X such that Xn =∼ X. However, with this particular construction the resulting isomorphisms turn out to be trivial. That is, if one uses the proof of 5.1 to produce an order X isomorphic to X3, and then analyzes the isomorphism f : X3 → X yielded by the proof and ω the associated decomposition X (I[u]2 ), one finds that I[u]2 = 1 if and only if u is eventually constant. Otherwise I[u]2 = ∅. Since for eventually constant u we have 2 ∼ [u]2 = [au]2 = [u], such a decomposition already witnesses X = X without the need for a p.r.a. of Xω. We shall now show how to directly build an order X along with an isomorphism ω F of X onto an order of the form X (I[u]). More generally we can construct X ω and F with F : X → X (I[u]n ). It will follow from the construction that these decompositions can be arranged to be non-trivial. We have observed that if the embedding from Xi × Yi into Xi+1 × Yi+1 is of the form (fi, gi), we have that the limit of the products Xi × Yi is isomorphic to the product of the limits. An analogous fact holds for limits of replacements, as well as for limits of infinite products. Let us spell these out. Suppose we are given a system

f0 f1 X0 −→ X1 −→ ... Let X = lim X . We again view each X as included in all subsequent X as well −→ i i j as in X. Suppose further that for every x ∈ X we are given a system

g0,x g1,x I0,x −−→ I1,x −−→ ... For each x ∈ X, let I = lim I . x −→ i,x Naturally we obtain a system

F0 F1 X0(I0,x) −→ X1(I1,x) −→ ...

The embedding Fi is deﬁned by Fi(x, y) = (fi(x), gi,x(y)), and in the replacement Xi(Ii,x), it is understood that the index x of each Ii,x only ranges over Xi. If we let Y = lim X (I ), then one may verify that Y ∼ X(I ). −→ i i,x = x Similarly, if we are given f0 f1 X0 −→ X1 −→ ... then we obtain a system ω F0 ω F1 X0 −→ X1 −→ ... where the embedding Fi is deﬁned by Fi((x0, x1,...)) = (fi(x0), fi(x1),...). We use the notation Fi = (fi, fi,...). Viewing each Xi as included in Xi+1 by way of f , we may view Xω as included in Xω by way of F . Letting Y = lim Xω, i i i+1 i −→ i observe that it is not the case (naturally, at least) that Y is isomorphic to Xω, where X = lim X . Rather, Y is isomorphic to the subset of Xω consisting of −→ i sequences of “bounded rank,” that is, sequences (x0, x1,...) such that for some i, S ω for all n we have xn ∈ Xi. If the Fi are true inclusions then Y = i Xi . Note that Y is a union of tail-equivalence classes. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 29

Theorem 5.2. Let {Lj : j ∈ J} by any collection of nonempty, pairwise non- isomorphic orders, indexed by some indexing set J. Then there exists an X such ∼ ω 2 ∼ that X = X (I[u]) for some collection of orders I[u] (and hence X = X), and further such that ω 1. For u, v ∈ X , if [u] 6= [v] then either I[u] = I[v] = ∅ or I[u] and I[v] are non-isomorphic. 2. For every j ∈ J, there exists a unique tail-equivalence class [u] such that I[u] = Lj. ω Proof. Let X0 be any order such that the number of tail-equivalence classes in X0 is at least |J|. ω Fix an injection ι : J → {[u]: u ∈ X0 }. If ι(j) = [u], define I0,[u] = Lj. For those [u] not assigned an index j ∈ J, pick orders I0,[u] so that the final collection ω {I0,[u] : u ∈ X0 } consists of pairwise non-isomorphic orders. ω Denote the order X0 (I0,[u]) by X1. Fix an embedding f0 : X0 → X1. For example, we may define f0 by f0(x) = (x, x, x, . . . , ax), where ax is any element in I0,[x]. By way of this f0, view X0 as included in X1. Then as in the discussion preceding ω ω the theorem, we may view X0 as included in X1 by way of F0 = (f0, f0,...). For ω ω each u ∈ X1 , if u ∈ X0 , define I1,[u] = I0,[u]. The remaining u are exactly those sequences with infinitely many terms from X1 \X0. For these [u], iteratively choose ω orders I1,[u] so that the final collection {I1,[u] : u ∈ X1 } consists of pairwise non- isomorphic orders. ω ω ω Let X2 = X1 (I1,[u]). The “inclusion” F0 : X0 → X1 naturally determines ω an “inclusion” f1 : X1 → X2. Namely, if (u, a) ∈ X1 = X0 (I0,[u]), with say u = (u0, u1,...), we define f1((u, a)) = (F0(u), a) = (f0(u0), f0(u1), . . . , a). It is legal to let f1 be the identity on the last coordinate, since the interval (u, ·) in X1 is of type I0,[u], and the interval (F0(u), ·) in X2 is of type I1,[u] = I0,[u]. ω Now repeat this process. View X1 as included (by way of f1) in X2, and X1 as ω ω ω included in X2 . For each u ∈ X2 , if u ∈ X1 , let I2,[u] = I1,[u]. For the remaining ω u, fix orders I2,[u] so that the collection {I2,[u] : u ∈ X2 } consists of pairwise non-isomorphic orders. Continuing in this way, we get a system

f0 f1 f2 X0 −→ X1 −→ X2 −→ ... Let X be the limit. Then we have

X = lim Xi ∼ −→ = lim Xi+1 −→ ω = lim Xi (Ii,[u]) −→ω = X (I[u]) ω ω where if u ∈ Xi for some i, then I[u] = Ii,[u], and if u is in none of the Xi (i.e. if u has terms of unboundedly high rank), then I[u] = ∅. This X satisfies the conclusion of the theorem. In particular, any fixed order L can appear as an interval in the X constructed in the proof, by simply including L among the Lj. By an analogous construction, for any n ≥ 1 we can get an X such that X =∼ ω n+1 ∼ X (I[u]n ) (and hence X = X), where the nonempty I[u]n are pairwise non- ω isomorphic and fill up the classes of every Xi for some increasing sequence X0 ⊆ 30 GARRETT ERVIN

X1 ⊆ ... converging to X. Although not all of the n-tail-equivalence classes [u]n ω are filled in the final decomposition X (I[u]n ), those that are filled are filled with pairwise non-isomorphic orders. In particular, there will be many examples of where ∼ I[u]n =6 I[au]n , and therefore these decompositions will be non-trivial. By doing a ∼ ω longer induction, it is possible to get X = X (I[u]n ) where every I[u]n is nonempty. Of course, it is also possible to arrange during the construction that some (or all) of the I[u]n are isomorphic.

6. Constructing A and X such that X =∼ A2X =6∼ AX Our main theorem gives that the cube property holds for the class of linear orders under the lexicographical product. In view of the proof, it is natural to ask if we could have established the stronger result, that for all orders A and X, if A2X =∼ X then AX =∼ X. By 3.13, if it were possible to construct a p.r.a. for any order of the form Aω then the answer would be yes. This raises the subquestion of whether constructing a p.r.a. for Aω is always possible. We show in this section that the answer to both questions is no. In fact, if A is a complete (κ, λ)-order for cardinals κ and λ of uncountable cofinality, then Aω does not admit a p.r.a. Furthermore, the converse to Theorem 3.13 holds: if Aω does not have a p.r.a., then there exists an X such that A2X =∼ X but AX =6∼ X. ∗ Thus in particular, there is such an X when A = ω1 + ω1. In the language of the introduction, this says that the (left-sided) weak Schroeder-Bernstein property fails for the class of linear orders. (The right-sided weak Schroeder-Bernstein property also fails, that is, there exist X,A with XA2 =∼ X but XA =6∼ X. This is easier to prove, but we will not do so here.) Given a linear order X, a cut in X is a pair of intervals (I,J) such that I∪J = X, I ∩ J = ∅, and I < J. Thus I is an initial segment of X, and J is a final segment. A cut (I,J) is called a gap if I has no maximal element, and J has no minimal element. The Dedekind completion of X, denoted X, is the order obtained from X by filling every gap (I,J) with a single point x(I,J). Viewing X as a suborder of X, we have that X is dense in X, and in particular the cofinality (respectively, coinitiality) of X coincides with that of X. The Dedekind completion X is always a complete linear order, and if X is complete to begin with, then X = X. Any automorphism f : X → X can be extended uniquely to an automorphism f : X → X. For, if (I,J) is a gap in X, then (f[I], f[J]) must also be a gap. The automorphism f is defined by setting f(x(I,J)) = x(f[I],f[J]) for every gap (I,J), and f(x) = f(x) for every x ∈ X. A subset C ⊆ X is called closed if every monotone sequence in C is either unbounded in X or converges to some point in C. For C to be closed it is necessary that C is complete in the order inherited from X. However, completeness is not 1 sufficient for closure. For example, { n : n ≥ 1} ∪ {0} is closed as a subset of R, 1 whereas { n : n ≥ 1} ∪ {−1} is not. Note that if X is not complete, then X is not closed as a subset of itself. A subset C ⊆ X is called left-unbounded if for every x ∈ X there exists c0 ∈ C such that c0 ≤ x, and right-unbounded if for every x one can find c1 ∈ C with c1 ≥ x. If C is unbounded in both directions, we simply say C is unbounded. If C is closed and left-unbounded, then C is called a left club, and C is a right club if it is closed and right-unbounded. If C is both a left and right club, then we say THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 31 simply that C is a club. Similarly, if I ⊆ X is an interval, then viewing I as an order in itself we may speak of a right club in I, left club in I, and club in I. It is straightforward to check that if X has uncountable cofinality, then the intersection of two right clubs is a right club, and if X has uncountable coinitiality, then the intersection of two left clubs is a left club. Hence if X has both uncountable cofinality and coinitiality, the intersection of two clubs is a club, the intersection of a club with a right club is a right club, and the intersection of a club with a left club is a left club. An order X may not contain any club suborders, but its completion X always contains at least one club, namely X itself. Suppose that X is a (κ, λ)-order, and both κ and λ have uncountable cofinality (so that X has uncountable cofinality and coinitiality). Let f be an automorphism of X, and f its extension to X. Let us check that C, the set of fixed points of f, is a club in X. It is clear that C is closed. To see that it is unbounded, fix x ∈ X. If f(x) = x, then x ∈ C and there is nothing to check. So suppose x 6∈ C. Then either f(x) > x or f(x) < x. Assume without loss of generality that we are in the former case. Then the positive iterates of x under f form an increasing sequence, that is, we have 2 x < f(x) < f (x) < . . . Since X has uncountable cofinality, this sequence is bounded, and hence converges (by completeness) to some point b. It is easy to see that b must be a fixed point of f. Symmetrically, since the coinitiality of X is also uncountable, the negative iterates of x converge to some a, and this a must be fixed by f. We have found a, b ∈ C with a < x < b, and so C is unbounded as claimed. ∗ Until further notice, let A denote the order ω1 + ω1. If α ∈ ω1 is an ordinal, we ∗ denote the corresponding element in ω1 by −α. We identify the 0 of ω1 with the 0 ∗ of ω1 . Thus A = . . . < −α < . . . < −1 < 0 < 1 < . . . < α < . . . ∗ ω Theorem 6.1. Let A = ω1 + ω1. Then A does not admit a parity-reversing automorphism. <ω ω Proof. For every finite sequence r ∈ A , let Ir denote the interval in A consisting of sequences beginning with r. We confuse sequences of length 1 with elements of A, so that if a ∈ A then Ia means the interval of points with first entry a. Note ω that Ir is isomorphic to A for every r, and in particular has neither a left nor right endpoint. ω While A is complete, A is not, since for every r the interval Ir has a gap to its immediate left and right. To see this, let J be the interval consisting of points in Ir and above, and let I be the complement of J. Then J has no minimum, since Ir does not. On the other side, I has no maximum, since any u ∈ I must begin with a finite sequence s, of the same length as r, but with some entry si < ri. Writing u = su0, pick a sequence u00 > u0, which is always possible since Aω has no top 00 point. Let v = su . Then v > u but v is still in I since it lies below Ir. Thus the cut (I,J) is in fact a gap, and it lies to the immediate left of Ir. Symmetrically ω (since A has no bottom point), Ir has a gap to the right. For every r ∈ A<ω, let r− and r+ denote the elements of Aω that fill the gaps to the left and right of Ir respectively. (These points are not pairwise distinct: if a, b ∈ A and b = a + 1 then for any r ∈ A<ω we have ra+ = rb−.) 32 GARRETT ERVIN

Let f : Aω → Aω be an automorphism of Aω, and f its extension to Aω. Let C ⊆ Aω denote the club of ﬁxed points of f. We shall show that f has a ﬁxed point of the form u = (α0, −α1, α2, −α3,...) for some collection of ordinals αi ∈ ω1. Since such a u cannot be periodic of odd period, we have that [u]2 6= [au]2, and thus f is not parity-reversing. Consider the ω1-length increasing sequence of intervals

I0 < I1 < . . . < Iα < . . . ω If α is a limit ordinal, then there are no points in A that lie below Iα but above each Iβ for β < α. This means that the ω1-sequence of left endpoints 0− < 1− < . . . < α− < . . . is closed in Aω. Since this sequence is right unbounded, it forms a right club in Aω. Denote this right club by D0. ω − Then D0 ∩ C is a right club in A . Fix α0 ∈ D0 ∩ C. Let J1 denote the interval ω − − of points in A lying strictly above α0 . Then since α0 is a fixed point of f, we have that f restricted to J1 is an automorphism of J1. The coinitiality of J1 is ω1 since Iα0 is an initial segment of J1, and its cofinality is also ω1, since it is a final ω segment of A . Hence J1 is a complete (ω1, ω1)-order. It follows that the set of fixed points of f in this interval, J1 ∩ C, is a club in J1. Now consider the descending sequence of intervals

I(α0,0) > I(α0,−1) > . . . > I(α0,−α) > . . . and the corresponding closed sequence of right endpoints + + + (α0, 0) > (α0, −1) > . . . > (α0, −α) > . . .

Denote this sequence by D1. Then D1 is left unbounded in J1, and hence a left + club in J1. Thus D1 ∩ C is a left club in J1, and we may ﬁx (α0, −α1) ∈ D1 ∩ C. At the next stage we deﬁne J2 to be the subinterval of J1 consisting of points + that lie strictly below (α0, −α1) . Then f restricted to J2 is an automorphism of

J2. This interval ends with I(α0,−α1) and hence has cofinality ω1. Since it also has coinitiality ω1, we have again that the set of fixed points in this interval, J2 ∩ C, − is a club in J2. Hence there must be a fixed point (α0, −α1, α2) in the right club sequence − − − (α0, −α1, 0) < (α0, −α1, 1) < . . . < (α0, −α1, α) < . . . − + Continuing in this way, we obtain a sequence of fixed points (α0) ,(α0, −α1) , − + ω (α0, −α1, α2) ,(α0, −α1, α2, −α3) ,... While these points all lie outside of A , ω they converge (in the obvious sense) to the point u = (α0, −α1, α2,...) ∈ A . Since the set of fixed points of f is closed, it must then be that u is fixed by f. Since f agrees with f on Aω, we have in fact that u is a fixed point of f. As observed already, it follows that f is not parity-reversing. It is easy to generalize the proof to get that if A is any complete (κ, λ)-order, where κ and λ have uncountable cofinality, then Aω does not have a parity-reversing automorphism. We will now prove the converse to 3.13. Recall that a linear order is called scattered if it does not contain an infinite, dense suborder. In particular, every ordinal α, considered as a linear order, is scattered. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 33

Theorem 6.2. Suppose that Aω does not have a parity-reversing automorphism. Then there exists an order X such that A2X =∼ X but AX =6∼ X. Proof. Since Aω does not have a p.r.a., it must be that A either has no endpoints, or only a single endpoint. In either case Aω is dense. Hence any interval in Aω, ω considered as a linear order itself, is dense. Suppose that A (Iu) is any replacement ω ω of A such that for densely many u we have Iu 6= ∅. Then since A is dense, for ω any interval I ⊆ A (Iu), we have that either I ⊆ Iu for some u, or I contains a dense suborder. (The “or” here is non-exclusive: both conditions may hold, but not neither.) In the latter case, I is by definition non-scattered. ω Let κ denote the number of ∼2-equivalence classes in A . Enumerate these ω classes as {Cα : α ∈ κ}. Let X be the replacement of A obtained by replacing ω every point in the αth class with a copy of α, that is, X = A (I[u]2 ) where I[u]2 = α if [u]2 = Cα. (It is inessential, though convenient, that we replace the points with ordinals. Any κ-sized collection of pairwise non-isomorphic scattered orders also works to define the I[u]2 .) By construction we have that A2X =∼ X. We know from 3.11 that AX =∼ ω A (J[u]2 ) where J[u]2 = I[au]2 for all u and a. Suppose toward a contradiction that ω ω there is an isomorphism f of X with AX, that is, of A (I[u]2 ) with A (J[u]2 ). For a fixed u, the order type of the interval Iu is an ordinal α, and in particular ω is scattered. Its image f[Iu] must be an interval of type α in A (J[u]2 ), and so from our observation above, it must be that f[Iu] ⊆ Jv for some v. By the same −1 −1 argument, f [Jv] must be contained in Iw for some w. But then since f [Jv] contains Iu, it must be that in fact v = w, and f[Iu] = Jv. But then Jv = α, and hence it must be that v ∈ [au]2. Moreover, f must be the identity on Iu, since the identity is the only isomorphism of α with itself. Thus the isomorphism f may be factored as (g, id), where g : Aω → Aω is a parity-reversing automorphism. But no such g exists, by hypothesis. Hence there is no isomorphism between X and AX.

7. Related Problems In Cardinal and Ordinal Numbers, Sierpińskiposes several other questions concerning the multiplication of linear orders aside from the cube problem. On page 232 he writes, “We do not know so far any example of two types ϕ and ψ, such that ϕ2 = ψ2 but ϕ3 6= ψ3, or types γ and δ such that γ2 6= δ2 but γ3 = δ3.” Later, on page 251, “We do not know whether there exist two different denumerable order types which are left-hand divisors of each other. Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other.” Since Sierpińskiordered products anti-lexicographically, “left-hand divisor” for him means “right-hand divisor” in the convention of this paper. Writing them out using our convention, his questions are, 1. Do there exist orders X and Y such that X2 =∼ Y 2 but X3 =6∼ Y 3? 2. Do there exist orders X and Y such that X2 =6∼ Y 2, but X3 =∼ Y 3? 3. Do there exist countable orders X and Y such that X =6∼ Y but for some orders A, B we have AY =∼ X and BX =∼ Y ? 4. Do there exist orders X and Y such that X =6∼ Y but for some orders ∼ ∼ ∼ ∼ A0,A1,B0,B1 we have A0Y = YA1 = X and B0X = XB1 = Y ? 34 GARRETT ERVIN

In comparison with the questions from the introduction, these questions are phrased negatively, asking for counterexamples to the corresponding properties. Sierpińskiwas aware of counterexamples to the unique square root property for linear orders, that is, of non-isomorphic orders X and Y such that X2 =∼ Y 2. These examples are due to Morel; see [10]. Question 2 is the natural generalization of the unique square root problem, and could be called the unique cube root problem. More generally, one may ask if there exist orders X and Y such that Xn =∼ Y n but Xk =6∼ Y k for k < n. Question 1 is motivated by the fact that Morel’s examples X,Y of non-isomorphic orders with isomorphic squares have the property that Xn =∼ X for all n ≥ 1 and Y n =∼ X for all n > 1. In particular, not only is it the case that X2 =∼ Y 2 but actually that Xn =∼ Y n for all n > 1. Question 1 asks whether this kind of collapsing is necessary, or if it is possible that two orders have isomorphic squares but non-isomorphic cubes. Both Questions 1 and 2 are related to a generalization of the cube problem. Suppose it were possible to find an order X such that X5 =∼ X but the powers X,X2,X3, and X4 were pairwise non-isomorphic. Then X and Y = X3 would give a positive answer to Question 2. Similarly, if it were possible to find an X isomorphic to X7 but whose intermediate powers were pairwise non-isomorphic, then X and Y = X3 would give a positive answer to Question 1. By our main theorem, there are no such X, but it may still be that these two questions have positive answers. These questions are, to the author’s knowledge, still open. Question 4 might be called the two-sided Schroeder-Bernstein problem for the class (LO, ×). It is a sensible problem to ask given that the lexicographical product is non-commutative and that there exist examples witnessing the failure of left- sided Schroeder-Bernstein property and right-sided Schroeder-Bernstein property for linear orders. It is closely related to the cube problem, in the sense that if there existed an X isomorphic to its cube but not its square, then X and Y = X2 would give a positive answer. There is no such X, but it turns out that Question 4 still has a positive answer. The proof of this will appear separately. This gives further evidence that the cube property for (LO, ×) is “close” to being false. Question 3 is the left-sided Schroeder-Bernstein problem for countable linear orders. Sierpińskiwas aware of uncountable orders X, Y, A, B satisfying the relations in the problem. In Section 6 we constructed such orders with A = B. From the discussion following 4.5, we know that in the case when A = B, any orders X and Y satisfying the relations of Question 3 must be uncountable. We will now strengthen this result, and show that Question 3 has a negative answer. Suppose that X, Y, A, B are countable orders, and AY =∼ X and BX =∼ Y . We prove X =∼ Y . Notice first that the hypotheses give ABX =∼ X. Let C = AB. There are three cases. If C has both endpoints, then both A and B must also have both endpoints. But then X, which is isomorphic to AY , contains an initial copy of Y , and Y , which is isomorphic to BX, contains a final copy of X. Hence X =∼ Y by Lindenbaum’s theorem. If C has neither endpoint, then by the proof of 3.19, since ∼ CX = X we have that X must be of the form Q(Ii). Thus X is invariant under left multiplication by any countable order. In particular, BX =∼ X, that is, Y =∼ X. Finally, suppose C has a single endpoint. Without loss of generality, assume it is the left endpoint. Then both A and B have a left endpoint (and at least one of them is missing the right endpoint). By the adaption of the proof of 3.19 to the left THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 35 endpoint case, X must be of the form Q(Ii), where Q = Q ∩ [0, 1), and where the order replacing the left endpoint 0 also has a left endpoint. It can be shown that such an order is invariant under left multiplication by any countable order with a left endpoint. In particular, BX =∼ X, that is, Y =∼ X. All of Sierpiński’s questions are instances of a much more general question. If we distinguish the structures in a given class C only up to isomorphism type, then (C, ×) may be viewed as a (possibly very large) semigroup, where the semigroup operation is given by the product. For a given semigroup (S, ·), we say that S can be represented in C if there is a map ι : S → C such that ι(a · b) =∼ ι(a) × ι(b), and if a 6= b then ι(a) =6∼ ι(b). The failure of the cube property for a given class C is equivalent to the statement that the group Z2 can be represented in C. Thus our main theorem gives that Z2, and more generally Zn, cannot be represented in (LO, ×). 5. Which semigroups can be represented in (LO, ×)? 6. Can any nontrivial group be represented in (LO, ×)? Question 6 is of course a subquestion of Question 5. By our results, if Question 6 has a positive answer, then any non-identity element in the witnessing group must be of infinite order. Questions 1, 2, and 4 all concern relations that can be realized in certain semigroups. Thus a complete answer to Question 5 would yield answers to all of these questions.

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